Take a breath before you start this one. Current Electricity has a reputation for being the chapter where the numericals bite, and I want to tell you honestly why that reputation exists: it is not because the physics is hard, it is because this chapter asks you to hold two pictures in your head at once. There is the microscopic picture, where you are tracking one electron staggering through a lattice of metal ions, and there is the circuit picture, where you are counting volts and amperes at terminals. Students who keep those two pictures separate find this chapter confusing. Students who learn to slide between them find it easy. That sliding is exactly what we are going to practise together.
To make the sliding automatic, this whole chapter is built on one carried-through analogy that I will call the Pump-and-Pipe Ledger. Water in pipes, a pump that lifts it, friction inside the pump itself, and a ledger that must always balance. Every single idea in this chapter — drift velocity, resistivity, EMF, internal resistance, Kirchhoff, even the Wheatstone bridge — will be hung on that one hook. You will also meet a sign-convention memory device of my own, “Rivers Down, Ladders Up”, which kills the single most common source of lost marks in Kirchhoff problems. By the end you will have worked through twenty solved problems, a full current electricity formula list class 12 in one table, and a practice worksheet built from the kind of current electricity class 12 important questions that actually turn up.
One prerequisite check before we begin. This chapter assumes you are comfortable with electric field and potential difference from Electric Charges and Fields, the Class 12 Physics chapter that sets up field and force. If the phrase “potential difference” still feels vague to you, spend twenty minutes there first. Everything below will land better.
What You’ll Learn
- Electric Current: What Actually Flows in a Metal
- Drift Velocity, Relaxation Time and Mobility
- Ohm’s Law and V–I Characteristics (Linear and Non-Linear)
- Electrical Resistivity and Conductivity
- Temperature Dependence of Resistance
- Electrical Energy and Power
- Resistors in Series and Parallel: Your Working Tool
- EMF, Terminal Voltage and Internal Resistance of a Cell
- Combination of Cells in Series and in Parallel
- Kirchhoff’s Rules Solved Examples Class 12
- Wheatstone Bridge and the Balance Condition
- Current Electricity Formula List Class 12
- Current Electricity Important Questions: The Traps That Cost Marks
- Practice Worksheet: Current Electricity Numericals With Solutions
Your Game Plan
Do not read this chapter front to back in one sitting. Work it in this order instead, because each step is the foundation of the next:
- Build the microscopic picture first. Current, drift velocity, mobility. Two days. Do not rush past this — it is where the “why” of every later formula lives.
- Then the material properties. Ohm’s law, resistivity and conductivity, temperature dependence. One day. Learn to say out loud what resistance depends on and what it does not depend on.
- Then the energy accounting. Power, heating, kilowatt-hours. Half a day. Cheap marks live here.
- Then the cell. EMF versus terminal voltage, internal resistance, combinations. One day. This is where most students quietly go wrong, so go slow.
- Then the circuit machinery. Series and parallel resistors as a tool, then Kirchhoff, then Wheatstone. Two days, and make every one of them a pen-on-paper day.
- Then the worksheet at the bottom with a closed book and a clock. Then re-read only the sections where you slipped.
Study Notes
Electric Current: What Actually Flows in a Metal
Start with the definition, which is deliberately simple. Pick any cross-section of a wire — a flat imaginary slice straight through it. Watch it for a time interval and count the net charge that crosses. Electric current is that charge divided by that time.
If the flow is steady, I = q / t. If the flow is changing from moment to moment, you want the instantaneous value, which is the derivative: i = dq/dt. The SI unit is the ampere, and one ampere means one coulomb of charge crossing your slice every second. Current is a scalar, even though we draw arrows for it — those arrows only record direction along a wire, and currents at a junction add like ordinary numbers, not like vectors. That is a favourite one-mark question.
Now the part that trips people. Inside a metal, the things that actually move are free electrons, and they are negative. So the electrons drift one way and we say the current points the other way. This is not a mistake anybody is hiding from you; it is a historical convention fixed before the electron was discovered, and physics simply kept it. Conventional current is in the direction positive charge would move, which is opposite to electron drift in a metal.
Where do those free electrons come from? In a metal such as copper, each atom releases roughly one outer electron into a shared pool. What is left behind is a lattice of positive ions, locked in place, vibrating about fixed positions. The freed electrons wander through this lattice at enormous speeds — of the order of 105 m/s at room temperature — but completely at random, colliding constantly with the vibrating ions. Random means no net transport: as many electrons cross your slice leftward as rightward, so with no battery connected the current is exactly zero even though every electron is moving fast.
Connect a battery and you superimpose a tiny, systematic bias on that chaos. That bias is the current. In Pump-and-Pipe language: the water molecules in a still pipe are already jiggling violently in every direction; switching on the pump adds a slow, common drift on top of the jiggling. The jiggling was always there and carries nothing. The drift is what fills your bucket.
Step 1 — total charge. q = ne × e = (4.8 × 1019)(1.6 × 10−19 C) = 7.68 C.
Step 2 — divide by time. The interval is one minute = 60 s.
I = q / t = 7.68 / 60 = 0.128 A (that is 128 mA).
Notice the electron count is huge and the current is modest. That ratio — roughly 6.25 × 1018 electrons per coulomb — is worth memorising as a sanity check.
(a) Instantaneous. i = dq/dt = 6t + 5. At t = 2: i = 6(2) + 5 = 17 A.
(b) Average. Average current is total charge divided by total time, not the average of the instantaneous values.
q(2) = 3(4) + 5(2) + 2 = 24 C; q(0) = 2 C.
Δq = 24 − 2 = 22 C, so Iavg = 22 / 2 = 11 A.
Why it works: the derivative answers “how fast right now”; the difference quotient answers “how much in total, spread out”. Two different questions, two different tools. Mixing them up is a classic slip.
Drift Velocity, Relaxation Time and Mobility
Here is the picture, built one layer at a time. With no field, a free electron travels in a straight line, smacks into a vibrating ion, and ricochets off in a random new direction. The average time between two successive collisions is called the relaxation time, written τ (tau). In a typical metal it is around 10−14 seconds — unimaginably short.
Now switch on a field E along the wire. Between collisions, each electron feels a force F = −eE and therefore an acceleration a = −eE/m. It picks up a little extra velocity in the direction opposite to E. Then it collides, and the collision wipes the memory clean — the electron comes out in a random direction with no memory of the velocity it had gained. So the extra velocity never accumulates. It builds up for a time τ, gets reset, builds again, gets reset.
Averaged over all the electrons, the random parts cancel and only the field-driven part survives. That survivor is the drift velocity:
vd = − (eτ/m) E and in magnitude vd = eτE / m
Next, link drift velocity to current. Take a wire of cross-sectional area A with n free electrons per unit volume. In a time Δt, every electron moves a distance vdΔt along the wire, so every electron sitting inside a cylinder of length vdΔt and area A will make it across your slice. That cylinder has volume A vdΔt, so it holds n A vdΔt electrons, carrying charge n A vdΔt e. Divide by Δt:
I = n A e vd
That single equation is the bridge between the microscopic picture and the circuit picture, and it earns marks constantly. Divide both sides by A and you get current density, J = I/A = n e vd — the vector cousin of current, measured in A/m2 and pointing along E.
Finally, mobility. Mobility μ asks a fair question: for a given push, how responsive is this charge carrier? It is defined as drift speed per unit field:
μ = vd / E = eτ / m (units: m2 V−1 s−1)
Mobility is always taken as a positive quantity, and it is a property of the carrier in that material, not of the wire’s shape. Combining, I = n A e μ E, and J = n e μ E. Notice that mobility packages τ and m together, so a material whose electrons collide less often (larger τ) has higher mobility and conducts better. That is the whole story of conduction in one sentence.
Step 1 — rearrange. From I = nAevd, vd = I / (nAe).
Step 2 — build the denominator first. nAe = (8.5 × 1028)(1.0 × 10−6)(1.6 × 10−19) = 1.36 × 104.
Step 3 — divide. vd = 1.5 / (1.36 × 104) = 1.10 × 10−4 m/s.
About a tenth of a millimetre per second. Slower than a growing fingernail.
t = distance / vd = 1 / (1.103 × 10−4) = 9.07 × 103 s ≈ 2.5 hours.
The resolution. You are not waiting for that electron to arrive. The wire is already full of free electrons everywhere, including inside the bulb filament. When you close the switch, the electric field propagates along the conductor at close to the speed of light, and every electron everywhere — including the ones already sitting in the filament — starts drifting essentially at once. It is the pipe already being full of water: open the tap and water comes out immediately, not after the molecules from the reservoir have crawled the whole way.
From μ = eτ/m we get τ = μm / e.
τ = (4.5 × 10−3)(9.1 × 10−31) / (1.6 × 10−19)
τ = (4.095 × 10−33) / (1.6 × 10−19) = 2.56 × 10−14 s.
Cross-check with Example 3: if that same wire had a field of 0.02 V/m along it, the mobility would be μ = vd/E = (1.103 × 10−4)/0.02 = 5.51 × 10−3 m2 V−1 s−1 — the same order of magnitude, which tells you the numbers in these problems are realistic.
Ohm’s Law and V–I Characteristics (Linear and Non-Linear)
Ohm’s law is an experimental statement, not a law of nature that everything must obey. It says: for a given conductor at constant physical conditions, the current through it is directly proportional to the potential difference across it. That last clause — at constant physical conditions, meaning chiefly constant temperature — is the part students drop and examiners look for.
Written out: V ∝ I, so V = IR, where the constant R is the resistance, measured in ohms (Ω). One ohm is one volt per ampere. Resistance is the pipe’s opposition to flow: how hard the pump must push to get one litre per second through.
Ohm’s law also has a beautiful microscopic form. Take vd = eτE/m and put it into I = nAevd:
I = nAe × (eτE/m) = (ne2τ/m) A E → J = (ne2τ/m) E = σE
and if the wire has length L with a uniform field, E = V/L, giving
R = mL / (ne2τA)
Look at what just happened. Ohm’s law fell out of the collision picture, and the constant R turned out to depend on the material (through n and τ) and on the geometry (through L and A) — and on nothing else. That is why R is a constant for a given wire at a given temperature, and it is also the clue to why heating breaks Ohm’s law: heating changes τ.
Now the graphs, because “draw the V–I characteristic” is an almost guaranteed question. Plot I vertically against V horizontally.
- Ohmic (linear). A metallic conductor at constant temperature, a manganin or nichrome wire, a standard resistor. The graph is a straight line through the origin. Its slope is 1/R, so a steeper line means a smaller resistance.
- Non-ohmic (non-linear). A filament lamp, a semiconductor diode, a thyristor, an electrolyte, a vacuum tube. The graph is curved, or not through the origin, or different in the two directions. Resistance is then not a single number — you must quote it at a stated operating point as V/I there.
The filament lamp in the diagram above is the cleanest example to describe in an exam. At low voltage the filament is cool, its resistance is low, and the curve rises steeply. As you raise the voltage the filament gets hotter, its resistance rises, and the curve bends over towards the voltage axis. It is still perfectly good physics — it just is not Ohm’s law, because the “constant physical conditions” clause has been violated.
Component A: (2 V, 0.50 A), (4 V, 1.00 A), (6 V, 1.50 A), (8 V, 2.00 A).
V/I = 4.00, 4.00, 4.00, 4.00 Ω. Constant → ohmic, R = 4 Ω.
Component B: (2 V, 0.61 A), (4 V, 0.98 A), (6 V, 1.31 A), (8 V, 1.60 A).
V/I = 3.28, 4.08, 4.58, 5.00 Ω. Rising → non-ohmic, and the rise tells you the component is heating up, so it is a filament lamp rather than a diode (a diode would also block reverse current).
Method to remember: do not eyeball the graph, compute V/I row by row. If the column of ratios is constant, it is ohmic. That one line of working is usually worth the mark on its own.
Electrical Resistivity and Conductivity
Resistance is a property of a particular object. Cut a wire in half and its resistance halves — but nothing about the copper changed. So we need a quantity that describes the material alone. That is resistivity, ρ.
Experiment gives two clean proportionalities for a uniform wire: R ∝ L (longer pipe, more opposition) and R ∝ 1/A (fatter pipe, less opposition). Put them together with a constant of proportionality:
R = ρL / A so ρ = RA / L (units: Ω m)
Comparing with the microscopic result R = mL/(ne2τA) gives the deep version:
ρ = m / (ne2τ) and σ = 1/ρ = ne2τ/m = neμ
σ is conductivity, measured in siemens per metre (S m−1, equivalently Ω−1m−1). It is simply the reciprocal of resistivity, and the microscopic form of Ohm’s law is J = σE. Read that equation as: the current density a material produces is proportional to the field you apply, and σ is the constant of willingness.
Look hard at ρ = m/(ne2τ). It contains no L and no A. Resistivity genuinely does not care about the shape of the sample — only about how many carriers there are (n) and how freely they run between collisions (τ). That is the single most examinable sentence in this section. Rough orders of magnitude worth carrying: metals about 10−8 Ω m, semiconductors somewhere in the range 10−4 to 100 Ω m, and insulators up to 1016 Ω m or more — a spread of twenty-four orders of magnitude, the widest range of any physical property you will meet at school.
The same ρ idea shows up outside physics too: when you meet conductance of electrolyte solutions in the Class 12 Chemistry chapter on Solutions, the conductivity and molar conductivity defined there are the same σ = 1/ρ wearing a chemist’s coat. Recognising that saves you learning it twice.
Step 1 — area from diameter (halve it first!). r = 0.25 mm = 2.5 × 10−4 m.
A = πr2 = π(2.5 × 10−4)2 = π(6.25 × 10−8) = 1.9635 × 10−7 m2.
Step 2 — apply R = ρL/A.
R = (1.7 × 10−8)(20) / (1.9635 × 10−7) = (3.4 × 10−7) / (1.9635 × 10−7)
R = 1.73 Ω.
The most common error here is using the diameter as the radius, which makes A four times too big and R four times too small. Halve it. Every time.
The key physical fact: stretching does not create or destroy copper, so the volume is unchanged: A1L1 = A2L2. And ρ is unchanged, because it is still the same copper.
If L2 = 3L1, then A2 = A1L1/L2 = A1/3.
R2/R1 = (L2/A2) ÷ (L1/A1) = (3L1 ÷ (A1/3)) ÷ (L1/A1) = 9.
R2 = 9 × 5 = 45 Ω.
General result to memorise: stretch a wire to n times its length and R becomes n2R. Length went up by n and area came down by n, and the two effects multiply.
Resistance R does NOT depend on: the voltage applied, the current flowing, or the shape the wire is bent into. Coiling a wire up does not change its resistance one bit — L and A are both untouched.
Resistivity ρ depends on: material and temperature only. Not on L, not on A, not on mass.
Temperature Dependence of Resistance
Heat a metal and its resistance goes up. Heat a semiconductor and its resistance goes down. Both facts drop straight out of ρ = m/(ne2τ) once you ask which of n and τ changes.
In a metal: n is essentially fixed — every atom has already donated its electron and heating does not persuade it to donate more. But heating makes the lattice ions vibrate with larger amplitude, so a drifting electron collides more often. τ falls, and since ρ ∝ 1/τ, resistivity rises. In the pipe: the walls got rougher, so the same pump delivers less water.
In a semiconductor or an insulator: heating does liberate extra carriers — a lot of them, exponentially many. The increase in n overwhelms the decrease in τ, so ρ falls sharply. This is why a semiconductor has a negative temperature coefficient of resistance.
Over a modest temperature range a metal’s resistance is very nearly linear in temperature, which gives the working formula:
RT = R0 [ 1 + α(T − T0) ] and equivalently ρT = ρ0 [ 1 + αΔT ]
Here α is the temperature coefficient of resistance, with units per degree Celsius (or per kelvin — the size of a degree is the same in both, so α has the same numerical value either way). Rearranged, α = (RT − R0) / (R0ΔT), which reads as “fractional change in resistance per degree”. Always state the reference temperature T0 that your α belongs to, because α itself drifts slightly with the choice.
ΔT = 120 − 20 = 100 °C.
R = 6.0 [1 + (4.0 × 10−3)(100)] = 6.0 [1 + 0.40] = 6.0 × 1.40 = 8.4 Ω.
A 40% rise for a 100-degree rise. This is exactly why the filament lamp in the V–I graph earlier was non-ohmic — a lamp filament runs at well over 2000 °C.
ΔR = 6.2 − 5.0 = 1.2 Ω; ΔT = 227 − 27 = 200 °C; R0 = 5.0 Ω.
α = ΔR / (R0ΔT) = 1.2 / (5.0 × 200) = 1.2 / 1000 = 1.2 × 10−3 °C−1.
Why it works: α is defined as a fractional change, so you must divide by R0 and not just by ΔT. Forgetting the R0 gives 6 × 10−3 with wrong units and loses the mark.
Electrical Energy and Power
Push a charge q through a potential difference V and you do work qV on it. In a resistor that work does not end up as kinetic energy of the electrons — the drift speed stays constant — it ends up shaking the lattice, which is to say it ends up as heat. That is the whole energy story of a resistor in one sentence.
Work done in time t when a steady current I flows: W = qV = (It)V. Divide by t to get the rate, which is power:
P = VI = I2R = V2/R
The last two forms come from substituting V = IR. All three are the same statement, and choosing the right one is most of the skill. The rule I want you to internalise: use the form built from the quantity that is the same for the components you are comparing. In a series circuit the current is common, so P = I2R is the clean form and the biggest resistance dissipates the most power. In a parallel circuit the voltage is common, so P = V2/R is the clean form and the smallest resistance dissipates the most power. Two sentences, and a whole family of bulb questions collapses.
The heat produced in time t is H = I2Rt, which is Joule’s law of heating. In practical life we bill energy in kilowatt-hours: 1 kWh is the energy used by a 1 kW appliance running for 1 hour, and 1 kWh = 3.6 × 106 J. On an electricity bill one kWh is called one “unit”.
One more idea you must be able to state: when an appliance is stamped “220 V, 1100 W”, those are the rated values — the appliance draws 1100 W only when exactly 220 V is across it. Its resistance is fixed at R = Vrated2/Prated; the power it actually consumes changes the moment the applied voltage changes.
(a) P = VI → I = P/V = 1100/220 = 5.0 A.
(b) R = V/I = 220/5.0 = 44 Ω. (Cross-check: R = V2/P = 48400/1100 = 44 Ω. Same answer, two routes — always worth doing.)
(c) Energy = 1.1 kW × 2.5 h = 2.75 kWh = 2.75 × 3.6 × 106 = 9.9 × 106 J.
(d) Cost = 2.75 × 7.00 = ₹19.25.
Tip: for part (c), keep power in kilowatts and time in hours and the answer comes out in kWh with no conversion at all.
Step 1 — resistances from the ratings. R = V2/P.
R60 = 48400/60 = 806.67 Ω R100 = 48400/100 = 484 Ω.
Step 2 — series total and current.
Rtotal = 806.67 + 484 = 1290.67 Ω; I = 220/1290.67 = 0.1705 A.
Step 3 — power in each, using the common quantity (current).
P60 = I2R60 = (0.1705)2(806.67) = 23.44 W
P100 = I2R100 = (0.1705)2(484) = 14.06 W
Answer: the 60 W bulb glows brighter. The bulb with the lower rating has the higher resistance, and in series higher resistance means more power.
Check the books balance: 23.44 + 14.06 = 37.5 W, and directly Ptotal = V2/Rtotal = 48400/1290.67 = 37.5 W. The ledger closes. (In parallel the verdict flips: there the 100 W bulb is brighter, because voltage is then the shared quantity.)
Resistors in Series and Parallel: Your Working Tool
A short, honest note first: series and parallel combination of resistors is not separately listed as a topic in your 2026-27 syllabus. It is listed machinery — you simply cannot do Kirchhoff’s rules or the Wheatstone bridge without it, and every circuit numerical assumes it. So treat this section as a tool you sharpen, not a topic you revise.
Series means end to end, one single path. The same current must pass through every component (there is nowhere else for it to go), and the potential differences add up to the supply voltage:
Rs = R1 + R2 + R3 + …
Parallel means side by side across the same two points. The same potential difference sits across every branch, and the branch currents add up to the total:
1/Rp = 1/R1 + 1/R2 + 1/R3 + … (for just two: Rp = R1R2/(R1+R2))
(2) A parallel combination is always smaller than the smallest resistor in it.
If your answer for two resistors in parallel comes out bigger than either one, you forgot to invert at the end. In pipe language: two pipes in a row is a longer, harder path; two pipes side by side is a wider, easier path.
| Feature | Series | Parallel |
|---|---|---|
| What is common | Current I | Potential difference V |
| What adds up | Voltages: V = V1+V2 | Currents: I = I1+I2 |
| Equivalent resistance | Rs = R1+R2 | Rp = R1R2/(R1+R2) |
| Compared to the parts | Bigger than the biggest | Smaller than the smallest |
| Best power formula | P = I2R → biggest R gets hottest | P = V2/R → smallest R gets hottest |
| If one component fails open | Everything stops | The others keep working |
Step 1 — collapse the parallel pair. Rp = (6 × 3)/(6 + 3) = 18/9 = 2 Ω. (Sanity check: 2 Ω is smaller than 3 Ω. Good.)
Step 2 — add the series resistor. Rtotal = 2 + 4 = 6 Ω.
Step 3 — total current. I = V/R = 12/6 = 2 A.
Step 4 — expand back out. All 2 A passes through the 4 Ω, dropping 8 V. So the parallel pair has 12 − 8 = 4 V across it.
I6 = 4/6 = 0.67 A I3 = 4/3 = 1.33 A
Step 5 — check. 0.67 + 1.33 = 2.00 A ✓ and the smaller resistor took the larger current, as it must.
The method — collapse inward to get the total, then expand back outward to get the details — solves almost every network question you will be set.
EMF, Terminal Voltage and Internal Resistance of a Cell
This is the section where careful students quietly pull ahead, so slow down here.
A cell is a chemical pump. Inside it, a chemical reaction drags positive charge from the low-potential plate to the high-potential plate — uphill, against the electric field. The work the cell does per unit charge in pushing charge around the complete circuit is called its electromotive force, ε:
ε = W / q (unit: volt)
Note immediately that despite the name, EMF is not a force. It is energy per unit charge, measured in volts. That misleading name is a nineteenth-century leftover, and examiners like asking about it.
Now the crucial complication. The electrolyte and electrodes inside the cell are not perfect conductors — charge has to fight its way through them too. That opposition is the cell’s internal resistance, r. In the Pump-and-Pipe Ledger, r is friction inside the pump’s own casing: the pump is rated to produce a certain pressure, but some of that pressure is spent shoving water through the pump itself, so less arrives at the outlet.
What a voltmeter across the terminals actually reads is the terminal potential difference V, and when the cell is discharging through an external resistance R:
ε = I(R + r) → I = ε/(R + r) and V = IR = ε − Ir
The quantity Ir is often called the lost volts, and that is a genuinely helpful name: it is the potential difference thrown away inside the cell, heating the cell instead of powering your circuit. Three consequences follow, and you should be able to recite them:
- While discharging, V < ε always. The terminal voltage is less than the EMF, and the gap widens as you draw more current.
- On open circuit, V = ε. No current flows, so Ir = 0 and the terminal voltage equals the EMF exactly. This is why an ideal voltmeter across an unconnected cell reads the EMF.
- While being charged, V > ε. The current is forced backwards through the cell, so V = ε + Ir. Keep this in your pocket — you will need it for the Kirchhoff problem below.
Two more useful rearrangements. From V = ε − Ir with I = V/R, you get r = (ε − V)/I and r = R(ε − V)/V — the second is the standard way to compute r when you know the external resistance and the terminal voltage.
Formula: V = ε − Ir when discharging, V = ε when I = 0, V = ε + Ir when charging.
Analogy: EMF is your salary; internal resistance is tax deducted at source; terminal voltage is what lands in your account. Draw more current and you climb into a higher tax bracket.
Current: I = ε/(R + r) = 2.0/(4.5 + 0.5) = 2.0/5.0 = 0.4 A.
Terminal voltage: V = IR = 0.4 × 4.5 = 1.8 V.
Lost volts: Ir = 0.4 × 0.5 = 0.2 V.
Check: 1.8 + 0.2 = 2.0 V = ε ✓
Energy audit for the same circuit: the cell supplies εI = 0.8 W; the external resistor gets VI = 0.72 W; the cell itself wastes I2r = 0.08 W as heat. The ledger balances to the last decimal.
Set up two equations in two unknowns using ε = I(R + r):
(i) ε = 0.50(4 + r) = 2 + 0.5r
(ii) ε = 0.25(9 + r) = 2.25 + 0.25r
Equate: 2 + 0.5r = 2.25 + 0.25r → 0.25r = 0.25 → r = 1 Ω.
Back-substitute: ε = 0.50(4 + 1) = 2.5 V.
Verify with (ii): 0.25(9 + 1) = 2.5 V ✓
Why it works: a single measurement gives one equation and two unknowns — underdetermined. The moment you change the external resistance you get an independent second equation, and the system closes. Any question that hands you two (I, R) pairs is asking for exactly this.
Combination of Cells in Series and in Parallel
One cell is rarely enough, so we group them. There are two arrangements and they solve two different problems.
Cells in series means the positive terminal of one joined to the negative terminal of the next, like several pumps in a row each adding pressure. For n cells of EMFs ε1, ε2, … and internal resistances r1, r2, …:
εeq = ε1 + ε2 + … req = r1 + r2 + …
and for n identical cells, εeq = nε and req = nr, giving I = nε/(R + nr). If a cell is connected the wrong way round, its EMF enters with a minus sign but its internal resistance still adds — resistance never cares about direction. That is a favourite trick.
Cells in parallel means all the positive terminals joined together and all the negative terminals joined together, like several pumps feeding one common manifold. For two cells:
εeq = (ε1r2 + ε2r1) / (r1 + r2) req = r1r2 / (r1 + r2)
and for n identical cells in parallel, εeq = ε (unchanged!) and req = r/n, giving I = ε/(R + r/n). Notice the equivalent EMF formula is a weighted average, with each cell weighted by the other cell’s internal resistance — the cell with the smaller internal resistance dominates, which is exactly what your intuition should say.
Use parallel when R is much smaller than r — then r/n dominates and dividing it by n multiplies your current. This is the arrangement for a low-resistance load, like a starter motor, which is why car batteries have thick, low-resistance internal construction.
Pipe version: pumps in a row give more pressure; pumps side by side give more flow.
εeq = 3 × 1.5 = 4.5 V; req = 3 × 0.4 = 1.2 Ω.
I = 4.5 / (1.2 + 3.8) = 4.5 / 5.0 = 0.9 A.
Vterminal = IR = 0.9 × 3.8 = 3.42 V.
Check: lost volts = I req = 0.9 × 1.2 = 1.08 V, and 3.42 + 1.08 = 4.50 V ✓
Equivalent EMF: εeq = (2.0 × 0.5 + 1.5 × 1.0)/(1.0 + 0.5) = (1.0 + 1.5)/1.5 = 2.5/1.5 = 1.67 V.
Equivalent internal resistance: req = (1.0 × 0.5)/1.5 = 0.5/1.5 = 0.33 Ω.
Current: I = 1.667/(3 + 0.333) = 1.667/3.333 = 0.5 A.
Terminal voltage: V = 0.5 × 3 = 1.5 V.
Independent check by junction currents. Let the common terminal voltage be V. Cell 1 supplies (2.0 − V)/1.0 and cell 2 supplies (1.5 − V)/0.5, and together they must equal V/3:
(2.0 − V) + 2(1.5 − V) = V/3 → 5 − 3V = V/3 → 15 − 9V = V → V = 1.5 V ✓
Cell 1 supplies 0.5 A and cell 2 supplies exactly 0 A — its EMF happens to equal the terminal voltage, so it is idling. A lovely reminder that a cell in a network does not have to be delivering anything.
Kirchhoff’s Rules Solved Examples Class 12
Series and parallel will take you a long way, but plenty of real circuits cannot be collapsed that way — the moment you have cells sitting in more than one branch, there is no “series” and no “parallel” to find. Kirchhoff gave us two rules that handle absolutely any network, and they are nothing more than conservation of charge and conservation of energy in circuit clothing.
Kirchhoff’s Junction Rule (first rule). At any junction, the sum of currents entering equals the sum of currents leaving. Equivalently, taking currents in as positive and out as negative, ΣI = 0 at every junction. Basis: conservation of charge — charge does not pile up at a point in a steady circuit, so whatever arrives must leave.
Kirchhoff’s Loop Rule (second rule). Around any closed loop, the algebraic sum of all potential changes is zero: ΣΔV = 0, usually written Σε = ΣIR. Basis: conservation of energy — electric potential is single-valued, so if you walk right round a loop and come back to where you started, you must be back at the same potential.
All the difficulty in Kirchhoff problems is signs. So here is my memory device, and I want you to write it at the top of your rough page every single time.
RESISTORS ARE RIVERS. Current flows downhill through a resistor. So if you walk with the current (downstream), you go down: write −IR. If you walk against the current (upstream), you climb: write +IR.
CELLS ARE LADDERS. A cell lifts you from its short plate to its long plate. So if you enter at the − terminal and leave at the + terminal, you go up: write +ε. Enter at + and leave at −: write −ε.
Then set the total of your walk to zero. Notice that for a resistor the sign depends on the current direction, while for a cell it depends only on the terminals — never on which way the current happens to be going through it. Rivers care about the current; ladders do not.
And the reassurance that makes the whole thing safe: you may assume any current direction you like. If you guess wrong, the algebra hands you a negative number, which is the circuit politely telling you “that current is real and this big, but it flows the other way.” You never have to redo the problem. That single fact removes most of the anxiety students feel here.
Step 1 — assume directions. Let I₁ flow B→A through the left branch, I₂ flow B→A through the right branch, and I₃ flow A→B through the middle.
Step 2 — junction rule at A: current in = current out, so
(i) I₁ + I₂ = I₃
Step 3 — left loop, walking B → A up the left branch, then A → B down the middle.
Up the left branch: through the cell we enter at − and leave at + (LADDER UP) → +12; through the 2 Ω we walk with I₁ (RIVER DOWN) → −2I₁.
Down the middle: we walk with I₃ (RIVER DOWN) → −3I₃.
Total = 0: 12 − 2I₁ − 3I₃ = 0, so
(ii) 12 = 2I₁ + 3I₃
Step 4 — right loop, the same walk on the right side:
(iii) 4 = 2I₂ + 3I₃
Step 5 — solve. From (ii), I₁ = (12 − 3I₃)/2 = 6 − 1.5I₃. From (iii), I₂ = (4 − 3I₃)/2 = 2 − 1.5I₃.
Put both into (i): I₃ = (6 − 1.5I₃) + (2 − 1.5I₃) = 8 − 3I₃ → 4I₃ = 8 → I₃ = 2 A.
Then I₁ = 6 − 3 = 3 A and I₂ = 2 − 3 = −1 A.
Step 6 — read the minus sign. I₂ is negative, so 1 A really flows A→B through the right branch: current is being pushed into the 4 V cell’s + terminal. The 4 V cell is being charged by the 12 V cell.
Step 7 — three independent checks.
• Junction: 3 + (−1) = 2 ✓
• Potentials: VA − VB from the middle branch = 3 × 2 = 6 V. From the left branch = 12 − 2(3) = 6 V ✓. From the right branch, since it is being charged, = 4 + 2(1) = 6 V ✓. All three agree.
• Power: the 12 V cell delivers 12 × 3 = 36 W; the 4 V cell absorbs 4 × 1 = 4 W; heat dissipated = 32(2) + 12(2) + 22(3) = 18 + 2 + 12 = 32 W. And 36 − 4 = 32 W ✓. The ledger closes exactly.
ΣIin = ΣIout → 3 + 1.5 = 2 + I4 → I4 = 4.5 − 2 = 2.5 A, flowing out of the junction.
Why it works: no charge accumulates at a point in a steady circuit. If more charge arrived than left, the junction would build up charge and its potential would rise without limit — which does not happen.
Wheatstone Bridge and the Balance Condition
The Wheatstone bridge is the most elegant idea in this chapter, and its elegance is worth naming: it lets you measure a resistance without ever measuring a current or a voltage accurately. All you need is an instrument that can reliably tell you when something is zero. Null methods like this are far more accurate than reading a scale, because you are never fighting the instrument’s calibration.
The arrangement. Four resistances P, Q, R and S form a closed quadrilateral ABCD. A cell is connected across one diagonal (A to C) and a sensitive galvanometer across the other diagonal (B to D). P sits in arm AB, Q in BC, R in AD and S in DC. You adjust one of the resistances until the galvanometer shows no deflection at all. The bridge is then said to be balanced, and:
P / Q = R / S
The derivation, using Kirchhoff. At balance the galvanometer current IG = 0. That single fact does two things at once. First, the junction rule at B says the current arriving through P must all continue through Q, so one current I1 flows through both P and Q. Similarly one current I2 flows through both R and S. Second, no current through G means no potential drop across it, so VB = VD.
Now apply the loop rule to loop ABDA. Walking A→B we go downstream through P, so −I1P; walking B→D through the galvanometer contributes nothing since IG = 0; walking D→A we go upstream through R, so +I2R. Total zero:
−I1P + I2R = 0 → I1P = I2R …(1)
Apply the same walk to loop BCDB:
−I1Q + I2S = 0 → I1Q = I2S …(2)
Divide (1) by (2). The currents cancel completely — which is exactly the magic — and you are left with P/Q = R/S. So if you know three of the four resistances, the fourth follows: S = QR/P. Notice that the answer does not contain the EMF of the cell, its internal resistance, or the galvanometer’s resistance. Change the battery and the balance point does not move. That independence is why the method is so accurate.
Balance condition: P/Q = R/S → S = QR/P.
S = (8 × 6)/4 = 48/4 = 12 Ω.
Check: P/Q = 4/8 = 0.5 and R/S = 6/12 = 0.5 ✓ The ratios match, so the bridge really is balanced.
Getting the arms in the right places is half the battle. Write down which resistance is in which arm before you touch the formula, and remember the condition pairs the two arms that share the junction B with the two that share D.
Test: P/Q = 2/3 = 0.667 and R/S = 4/5 = 0.800. These are not equal, so the bridge is not balanced and current does flow through the galvanometer.
Required S: S = QR/P = (3 × 4)/2 = 6 Ω. So S must be increased from 5 Ω to 6 Ω.
Direction of deflection: since R/S is currently too large, the drop across R is larger than the balance case demands, which puts D at a lower potential than B. Current therefore flows through the galvanometer from B to D. (You are rarely asked for the direction, but you are often asked “will the galvanometer show a deflection?” — and the ratio test answers that in one line.)
Step 1 — test for balance first. Always.
P/Q = 10/20 = 0.5; R/S = 15/30 = 0.5. Balanced!
Step 2 — exploit it. At balance no current flows through the galvanometer arm, so you may simply delete that 25 Ω from the circuit. Its value never enters the answer.
Step 3 — the network is now just two series paths in parallel.
Path A–B–C: 10 + 20 = 30 Ω. Path A–D–C: 15 + 30 = 45 Ω.
Req = (30 × 45)/(30 + 45) = 1350/75 = 18 Ω.
Sanity check: 18 Ω is smaller than 30 Ω, the smaller of the two parallel paths. Good.
This is the single most useful exam application of the bridge. Whenever a network question shows you five resistors in a diamond-plus-bridge shape, test the ratio before you do anything else. If it balances, the middle resistor vanishes and a fearsome-looking problem becomes two lines of arithmetic.
Current Electricity Formula List Class 12
Here is the whole chapter compressed into one table. Copy it onto a single sheet by hand — the act of writing it out is worth more than reading it ten times — and keep that sheet for the night before the exam.
| Quantity | Formula | SI unit / note |
|---|---|---|
| Electric current | I = q/t i = dq/dt | ampere (A); scalar |
| Drift velocity | vd = eτE/m | m/s; about 10−4 m/s |
| Current from drift | I = nAevd | the microscopic–circuit bridge |
| Current density | J = I/A = nevd = σE | A/m2; a vector |
| Mobility | μ = vd/E = eτ/m | m2 V−1 s−1 |
| Ohm’s law | V = IR | R in ohm (Ω); constant T |
| Resistance (geometry) | R = ρL/A | stretch to n× length → n2R |
| Resistance (microscopic) | R = mL/(ne2τA) | shows where R comes from |
| Resistivity / conductivity | ρ = m/(ne2τ); σ = 1/ρ = neμ | Ω m; S m−1 |
| Temperature dependence | RT = R0[1 + αΔT] | α in °C−1; negative for semiconductors |
| Power | P = VI = I2R = V2/R | watt (W) |
| Joule heating / energy | H = I2Rt; 1 kWh = 3.6 × 106 J | joule (J); 1 kWh = 1 unit |
| Resistors in series / parallel | Rs = ΣR; 1/Rp = Σ(1/R) | series > largest; parallel < smallest |
| Cell in a circuit | ε = I(R + r); V = ε − Ir | V = ε + Ir while charging |
| Internal resistance | r = (ε − V)/I = R(ε − V)/V | ohm (Ω) |
| Cells in series | εeq = Σε; req = Σr | use when R >> r |
| Cells in parallel (two) | εeq = (ε1r2+ε2r1)/(r1+r2); req = r1r2/(r1+r2) | use when R << r |
| Kirchhoff’s rules | ΣI = 0 at a junction; ΣΔV = 0 round a loop | charge conservation; energy conservation |
| Wheatstone balance | P/Q = R/S, so S = QR/P | independent of ε, r and RG |
Current Electricity Important Questions: The Traps That Cost Marks
I have marked enough scripts to know that most lost marks in this chapter come from a short, repeating list. Read this section the morning of the exam.
- Radius versus diameter. Every wire question that gives a diameter is quietly testing whether you halve it. Halving errors change your answer by a factor of four.
- ε where V belongs. Ask each time: is this the rated push, or the reading across the terminals with current flowing?
- Slope of the I–V graph. With I on the vertical axis the slope is 1/R, not R. Label the axes, then write the sentence.
- Reversed cell in a series battery. Its EMF subtracts, but its internal resistance still adds.
- Forgetting to test the bridge for balance. Do the ratio test on any diamond-shaped network before you attempt anything clever.
- Wrong power formula. Series → P = I2R. Parallel → P = V2/R. Pick the formula built on the shared quantity.
- Units in temperature problems. α is a fractional change per degree, so you must divide by R0. And ΔT is the same number in °C or K — do not convert twice.
- Calling current a vector. It has direction but not vector addition. It is a scalar.
- Panicking at a negative current. A negative answer in Kirchhoff is information, not an error. State the magnitude and say the direction is opposite to the one you assumed.
- Writing dependent Kirchhoff equations. Count j − 1 junction equations and b − j + 1 loop equations, then stop.
One more habit, and it is the highest-return habit in physics: after every numerical, spend ten seconds asking whether the answer is reasonable. Is the parallel combination smaller than the smallest resistor? Is the terminal voltage less than the EMF? Does the power supplied equal the power dissipated? These checks catch nearly every arithmetic slip before the examiner does. If you want more practice at the habit of checking under time pressure, working through a full practice paper with answers is the fastest way to build it.
Practice Worksheet: Current Electricity Numericals With Solutions
Eleven original questions, closed book, pen and paper, forty minutes. Work each one fully before you open the answer — peeking early feels productive and teaches you nothing. Take e = 1.6 × 10−19 C throughout.
Q1. In 2.0 seconds, 6.25 × 1018 electrons cross a section of a wire. Find the charge transferred and the current.
Show Answer
I = q/t = 1.0/2.0 = 0.5 A.
Remember: 6.25 × 1018 electrons is exactly one coulomb. Worth memorising as a check.
Q2. A conductor of cross-sectional area 2.0 × 10−6 m2 carries 2.4 A. The free electron density is 7.5 × 1028 m−3. Find the drift speed.
Show Answer
nAe = (7.5 × 1028)(2.0 × 10−6)(1.6 × 10−19) = 2.4 × 104.
vd = 2.4 / (2.4 × 104) = 1.0 × 10−4 m/s.
Q3. A wire has resistance 4 Ω. It is stretched uniformly until its length is doubled. Find the new resistance, and explain in one line why it is not simply 8 Ω.
Show Answer
Because volume is conserved, doubling the length also halves the area. R = ρL/A therefore picks up a factor of 2 from L and a factor of 2 from 1/A — the effects multiply, giving 4, not 2.
Q4. A wire of resistivity 2.0 × 10−7 Ω m has length 5.0 m and uniform cross-sectional area 1.0 × 10−6 m2. Find its resistance and its conductivity.
Show Answer
σ = 1/ρ = 1/(2.0 × 10−7) = 5.0 × 106 S m−1.
Q5. A metal wire has resistance 10 Ω at 20 °C, with α = 5.0 × 10−3 °C−1 referred to 20 °C. Find its resistance at 100 °C. Would the answer be larger or smaller if the wire were a semiconductor?
Show Answer
R = 10[1 + (5.0 × 10−3)(80)] = 10[1 + 0.40] = 14 Ω.
For a semiconductor, α is negative, so heating would lower the resistance — heating liberates far more carriers (n rises steeply), and that outweighs the drop in relaxation time.
Q6. A bulb is rated 100 W, 250 V. Find its resistance and its rated current. If it is instead operated on a 200 V supply, what power does it consume (assume constant resistance)?
Show Answer
I = P/V = 100/250 = 0.4 A.
On 200 V: P = V2/R = 40000/625 = 64 W.
Note the power fell to 64% for a voltage drop of only 20% — power goes as the square of voltage.
Q7. A cell of EMF 6.0 V and internal resistance 0.5 Ω is connected to a 5.5 Ω resistor. Find the current, the terminal voltage, and the power wasted inside the cell.
Show Answer
V = IR = 1.0 × 5.5 = 5.5 V.
Power wasted inside = I2r = (1.0)2(0.5) = 0.5 W.
Check: total εI = 6.0 W, external VI = 5.5 W, difference 0.5 W ✓
Q8. Four identical cells, each of EMF 2.0 V and internal resistance 0.25 Ω, are connected in series across a 3.0 Ω resistor. Find the current, the terminal voltage of the battery, and the lost volts.
Show Answer
I = 8.0/(1.0 + 3.0) = 2.0 A.
V = IR = 2.0 × 3.0 = 6.0 V.
Lost volts = 8.0 − 6.0 = 2.0 V (= I req = 2.0 × 1.0 ✓).
Q9. A 10 V cell and a 4 V cell are connected in a single loop in opposition (their EMFs push in opposite directions) together with resistances of 2 Ω and 3 Ω. Both cells have negligible internal resistance. Find the current, and state which cell is being charged.
Show Answer
I = (10 − 4)/(2 + 3) = 6/5 = 1.2 A.
The current is driven by the stronger 10 V cell, so it is forced backwards through the 4 V cell. The 4 V cell is being charged.
Check the energy ledger: the 10 V cell delivers 10 × 1.2 = 12 W; the 4 V cell absorbs 4 × 1.2 = 4.8 W; the resistors dissipate (1.2)2(5) = 7.2 W. And 12 − 4.8 = 7.2 W ✓
Q10. In a Wheatstone bridge ABCD the arms are P = 4 Ω (AB), Q = 12 Ω (BC) and R = 6 Ω (AD), with the galvanometer between B and D. Find the value of S in arm DC for balance, and then the equivalent resistance between A and C.
Show Answer
Check: 4/12 = 0.333 and 6/18 = 0.333 ✓
At balance no current flows through G, so delete it. Two series paths in parallel:
A–B–C = 4 + 12 = 16 Ω; A–D–C = 6 + 18 = 24 Ω.
Req = (16 × 24)/(16 + 24) = 384/40 = 9.6 Ω.
Sanity check: 9.6 Ω < 16 Ω, the smaller path. Good.
Q11. The same 60 W and 100 W bulbs from Example 12 (both rated 220 V) are now connected in parallel across a 220 V supply. Which glows brighter, what is the total power drawn, and what is the combined resistance?
Show Answer
Total power = 60 + 100 = 160 W.
Combined resistance = V2/P = 48400/160 = 302.5 Ω.
(Check with resistances: 806.67 Ω and 484 Ω in parallel = (806.67 × 484)/(1290.67) = 302.5 Ω ✓)
Compare with Example 12: the same two bulbs, the same supply, but the verdict flips. In series the 60 W bulb wins (common current, P = I2R); in parallel the 100 W bulb wins (common voltage, P = V2/R). If you can explain that reversal out loud, you have understood this chapter’s power section completely.
Before you close this page, do one thing. Take a blank sheet and, without looking, draw the two-loop circuit from the Kirchhoff section and solve it. If you get I₁ = 3 A, I₂ = −1 A and I₃ = 2 A, you own this chapter. If you do not, you have found exactly the paragraph you need to reread, which is far more useful than a vague feeling that you should “revise current electricity”.
And keep the standard modest. You do not need to master this chapter tonight. Kaizen — aim for one more correct answer than yesterday. One extra numerical, one extra sign convention that finally clicks, one extra sanity check remembered. Do that daily and by February you will not recognise the student who opened this page.