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State Board · Class 12 · Physics

Lesson 9: Current Electricity

7 concepts · 56 practice questions with answers · Free to read

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Kirchhoff's current and voltage laws

From daily life

At a busy crossing as many vehicles leave as arrive — like currents at a junction.

Simple definition

Kirchhoff's rules handle complex networks where simple series-parallel reduction fails.

Kirchhoff's rules : junction and loopI₁I₂I₃ΣI = 0ΣV = 0

In the figure : ΣI = 0 at a junction; ΣV = 0 around a loop.

Rule 1: Current law (KCL) — At a junction, incoming currents = outgoing currents : ΣI = 0. It expresses conservation of charge.

  • Easy : Incoming 5 A; one outgoing 3 A ⇒ the other 2 A.
  • Medium : 2 A and 3 A come in, 1 A goes out ⇒ the fourth carries 4 A out.
  • Exam (CET) : Which conservation law at a junction? (1) charge (2) energy (3) momentum (4) mass → answer 1)

Rule 2: Voltage law (KVL) — The sum of potential changes around a closed loop is zero : ΣV = 0 (ΣE = ΣIR). It expresses conservation of energy.

  • Easy : 12 V cell; 2 Ω and 4 Ω in series ⇒ I = 12 ÷ 6 = 2 A.
  • Medium : Sign : walking along I gives −IR; going − to + through a cell gives +E.
  • Exam (CET) : KVL follows from? (1) energy conservation (2) charge (3) momentum (4) force → answer 1)

Rule 3: Sign convention — Choose the loop direction; a cell from − to + : +E; a resistor along I : −IR, against I : +IR. The sum is 0.

  • Easy : E = 10 V, R = 5 Ω : 10 − 5I = 0 ⇒ I = 2 A.
  • Medium : A negative answer ⇒ the actual direction is opposite.
  • Exam (CET) : If a cell is traversed + to −, E has sign? (1) − (2) + (3) 0 (4) unchanged → answer 1)

Derivation : Equation for one loop

  • Step 1 : Cell E with resistors R₁ and R₂ in series.
  • Step 2 : Loop direction = direction of I; cell : +E.
  • Step 3 : Resistors : −I R₁ − I R₂.
  • Step 4 : E − I(R₁ + R₂) = 0 ⇒ I = E ÷ (R₁ + R₂).
Remember

ΣI = 0 (junction); ΣV = 0 (loop).

Common mistake / CET trap

Sign errors are the most common mistake; fix the loop direction first.

Shortcut

Memory clue : "junction = charge, loop = energy".

Spot the difference

RuleStatementConservation
KCLΣI = 0charge
KVLΣV = 0energy

Practice questions

Easy
  1. 1What is KCL?
  2. 2What is KVL?
  3. 3KCL expresses which conservation?
Medium
  1. 46 A in; 4 A out; the third?
  2. 5E = 9 V, R = 3 Ω, 6 Ω : I?
  3. 6E = 6 V, r = 1 Ω, R = 5 Ω : I?
Challenging
  1. 7Cells 12 V and 4 V (opposing), R = 4 Ω : I?
  2. 8Currents 1, 2, 3 A enter a junction; how much leaves?
Show answers (try them yourself first)
  1. 1ΣI = 0
  2. 2ΣV = 0
  3. 3Charge
  4. 42 A
  5. 5I = 1 A
  6. 6I = 1 A
  7. 7I = 2 A
  8. 86 A

Kirchhoff's rules in networks and combinations of cells

From daily life

A torch uses 2–3 cells in series to get a larger voltage.

Simple definition

The emf is a cell's maximum potential; internal resistance r is its own resistance.

A two-loop networkI₁I₂I₁ − I₂E₁, E₂, R₁, R₂, R₃

In the figure : Two loops sharing a branch.

Rule 1: Two-loop network — Apply KVL to each loop and KCL at the junction; as many equations as unknown currents.

  • Easy : E₁ = 12 V, E₂ = 6 V (opposing), R₁ = 2, R₂ = 3, R₃ = 1 Ω : I₁ = 3.82 A, I₂ = -0.55 A.
  • Medium : Current in R₃ = I₁ − I₂ = 4.36 A.
  • Exam (CET) : 2 unknowns ⇒ equations needed? (1) 2 (2) 1 (3) 3 (4) 0 → answer 1)

Rule 2: Cells in series — E_eq = E₁ + E₂ + …, r_eq = r₁ + r₂ + …. For n identical cells : nE, nr.

  • Easy : 4 cells, E = 1.5 V, r = 0.5 Ω, R = 3 Ω : I = 1.2 A.
  • Medium : R ≫ nr ⇒ series is useful.
  • Exam (CET) : Series is best when? (1) R is large (2) R is small (3) never (4) r is large → answer 1)

Rule 3: Cells in parallel — n identical cells in parallel : E_eq = E, r_eq = r ÷ n. Useful when R is small.

  • Easy : 4 cells in parallel, E = 1.5, r = 0.5, R = 0.1 Ω : I = 1.5 ÷ (0.125 + 0.1) = 6.67 A.
  • Medium : E does not increase in parallel.
  • Exam (CET) : r_eq for parallel cells? (1) r/n (2) nr (3) r (4) 0 → answer 1)

Derivation : n cells in series : I = nE ÷ (R + nr)

  • Step 1 : Total emf = nE.
  • Step 2 : Total internal resistance = nr.
  • Step 3 : KVL : nE − I nr − I R = 0.
  • Step 4 : I = nE ÷ (R + nr).
Remember

Series nE, nr; parallel E, r/n; I = nE/(R + nr).

Common mistake / CET trap

Parallel does not raise emf; series raises r.

Shortcut

Memory clue : "series = force, parallel = lifetime".

Spot the difference

ConnectionE_eqr_eq
Series (n)nEnr
Parallel (n)Er/n

Practice questions

Easy
  1. 1E_eq in series?
  2. 2r_eq in parallel?
  3. 3When to use series?
Medium
  1. 45 cells, E = 2 V : E_eq?
  2. 53 cells in parallel r = 0.9 Ω : r_eq?
  3. 62 cells E = 1.5, r = 0.5, R = 2 Ω in series : I?
Challenging
  1. 76 identical cells : 2 groups of 3 in series, in parallel; E, r?
  2. 8How many equations for 3 branches, 2 loops?
Show answers (try them yourself first)
  1. 1E₁ + E₂ + …
  2. 2r/n
  3. 3When R is large
  4. 410 V
  5. 50.3 Ω
  6. 6I = 3 ÷ 3 = 1 A
  7. 7E_eq = 3E, r_eq = 1.5 r
  8. 82 loops + 1 junction = 3

Wheatstone bridge

From daily life

A precise balance equalises two pans — a bridge balances two resistance ratios.

Simple definition

A Wheatstone bridge measures an unknown resistance.

Wheatstone bridgePQRSGP/Q = R/S

In the figure : At balance G reads zero; P/Q = R/S.

Live figure

Wheatstone bridge

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Rule 1: Bridge principle — Four resistors P, Q, R, S and a galvanometer G. At balance no current flows through G : P ÷ Q = R ÷ S.

  • Easy : P = 10, Q = 20, R = 15 ⇒ S = 30 Ω.
  • Medium : At balance B and D are at equal potential.
  • Exam (CET) : Balance condition? (1) P/Q = R/S (2) PQ = RS (3) P = Q (4) R = S → answer 1)

Rule 2: Unbalanced bridge — When P/Q ≠ R/S current flows through G, in the direction given by the potential difference.

  • Easy : P = 6, Q = 6, R = 4, S = 8 ⇒ unbalanced (1 ≠ 0.5).
  • Medium : Balance persists if the battery is changed.
  • Exam (CET) : Current in G at balance? (1) 0 (2) maximum (3) infinite (4) reversed → answer 1)

Rule 3: Sensitivity — The bridge is most sensitive when all four resistances are of the same order.

  • Easy : P = Q = R = S = 100 Ω is best.
  • Medium : P = 1 Ω with S = 1 MΩ is poor.
  • Exam (CET) : Highest sensitivity? (1) same order (2) different (3) small R (4) large R → answer 1)

Derivation : Balance condition P/Q = R/S

  • Step 1 : At balance I_G = 0 ⇒ same I₁ in P and Q; I₂ in R and S.
  • Step 2 : B and D equipotential ⇒ I₁P = I₂R.
  • Step 3 : Similarly I₁Q = I₂S.
  • Step 4 : Dividing, P ÷ Q = R ÷ S.
Remember

P/Q = R/S ⇒ I_G = 0.

Common mistake / CET trap

Keep the positions of P, Q, R and S in order.

Shortcut

Memory clue : "ratios on both arms equal".

Spot the difference

StateI_GCondition
Balanced0P/Q = R/S
Unbalanced≠ 0P/Q ≠ R/S

Practice questions

Easy
  1. 1Balance condition?
  2. 2What is G?
  3. 3V at B and D when balanced?
Medium
  1. 4P = 5, Q = 10, R = 8 : S?
  2. 5P = 20, R = 30, S = 60 : Q?
  3. 6P/Q = 2 ⇒ R/S?
Challenging
  1. 7P = 12, Q = 4, R = 9 : S ?
  2. 8If battery and G are interchanged, balance?
Show answers (try them yourself first)
  1. 1P/Q = R/S
  2. 2Galvanometer
  3. 3Equal
  4. 4S = 16 Ω
  5. 5Q = 40 Ω
  6. 62
  7. 7S = 3 Ω
  8. 8Remains

Metre bridge

From daily life

Hold a rope in the middle and the two sides balance; a wire acts alike.

Simple definition

A metre bridge is a Wheatstone bridge made with a 1 m wire.

Metre bridgeRSGl(100 − l)R/S = l ÷ (100 − l)

In the figure : Null point at l; R/S = l/(100 − l).

Rule 1: Metre-bridge principle — A 1 m uniform wire forms two arms; at balance R ÷ S = l ÷ (100 − l).

  • Easy : R = 6 Ω, l = 40 cm ⇒ S = 9 Ω.
  • Medium : Uniform wire ⇒ resistance ∝ length.
  • Exam (CET) : l = 50 cm ⇒ R/S? (1) 1 (2) 2 (3) 0.5 (4) 0 → answer 1)

Rule 2: Specific resistance of the wire — S = ρL ÷ A = ρL ÷ (π d²/4) ⇒ ρ = S π d² ÷ 4L.

  • Easy : S = 4 Ω, d = 0.4 mm, L = 0.5 m : ρ = 1.01 × 10⁻⁶ Ω m.
  • Medium : Measure d with a screw gauge.
  • Exam (CET) : Unit of ρ? (1) Ω m (2) Ω (3) Ω/m (4) S → answer 1)

Rule 3: Accuracy and errors — Keep the null point near the middle (35–65 cm) by adjusting R. End resistances cause error.

  • Easy : l = 50 gives the best accuracy.
  • Medium : Heating changes ρ — pass current briefly.
  • Exam (CET) : Best position of l? (1) middle (2) near 0 (3) near 100 (4) anywhere → answer 1)

Derivation : R/S = l/(100 − l)

  • Step 1 : Two parts of the wire : resistance ∝ length; l and (100 − l).
  • Step 2 : Bridge balance : R ÷ S = (resistance of l) ÷ (resistance of 100 − l).
  • Step 3 : Wire uniform ⇒ ratio of lengths.
  • Step 4 : R ÷ S = l ÷ (100 − l).
Remember

S = R(100 − l)/l; ρ = Sπd²/4L.

Common mistake / CET trap

Do not swap l and 100 − l; R on the left ⇒ l measured from the left.

Shortcut

Memory clue : "R gets l, S gets the rest".

Spot the difference

QuantityFormulaUnit
SR(100 − l)/lΩ
ρSπd²/4LΩ m

Practice questions

Easy
  1. 1Metre-bridge principle?
  2. 2l = 50 ⇒ S?
  3. 3Where should the null point be?
Medium
  1. 4R = 8, l = 20 : S?
  2. 5R = 3, l = 60 : S?
  3. 6S = 2, d = 0.5 mm, L = 0.4 m : ρ?
Challenging
  1. 7R = 10, S = 5 : l?
  2. 8l changes 40 → 60 : change in S (R = 4)?
Show answers (try them yourself first)
  1. 1R/S = l/(100−l)
  2. 2S = R
  3. 3Near the middle
  4. 4S = 32 Ω
  5. 5S = 2 Ω
  6. 6ρ = 9.82 × 10⁻⁷ Ω m
  7. 7l = 66.7 cm
  8. 8S: 6 Ω → 2.67 Ω

Potentiometer principle and comparing emfs

From daily life

On a uniform slope the height drops equally per metre; along the wire the potential drops equally per cm.

Simple definition

A potentiometer measures emf using a uniform potential gradient.

Potentiometer : potential ∝ lengthElV = k l (k = V/L)

In the figure : V = k l; at the null point the cell's E = k l.

Live figure

Potentiometer principle and comparing emfs

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Rule 1: Principle of a potentiometer — Across a uniform wire carrying a steady current, potential difference ∝ length : V = k l; k = potential gradient (V/m).

  • Easy : Wire 4 m, 2 V ⇒ k = 0.5 V/m.
  • Medium : l = 3 m ⇒ V = 1.5 V.
  • Exam (CET) : k depends on? (1) I, ρ, A (2) only l (3) E (4) external R → answer 1)

Rule 2: Comparing emfs — E₁ ÷ E₂ = l₁ ÷ l₂ (the potentiometer draws no current).

  • Easy : E₁ = 1.2 V, l₁ = 60 cm, l₂ = 75 cm ⇒ E₂ = 1.5 V.
  • Medium : At the null point the cell gives no current.
  • Exam (CET) : Ratio of emfs? (1) l₁/l₂ (2) l₂/l₁ (3) l₁l₂ (4) 1 → answer 1)

Rule 3: Potentiometer vs voltmeter — The potentiometer has effectively infinite resistance at balance (zero current) ⇒ true emf; a voltmeter draws a little current ⇒ gives terminal voltage.

  • Easy : Voltmeter reads V = E − Ir.
  • Medium : A potentiometer gives E.
  • Exam (CET) : True emf with? (1) potentiometer (2) voltmeter (3) ammeter (4) galvanometer → answer 1)

Derivation : E₁/E₂ = l₁/l₂

  • Step 1 : On the wire V = k l.
  • Step 2 : First cell null at l₁ : E₁ = k l₁.
  • Step 3 : Second cell at l₂ : E₂ = k l₂.
  • Step 4 : E₁ ÷ E₂ = l₁ ÷ l₂.
Remember

V = kl; E₁/E₂ = l₁/l₂.

Common mistake / CET trap

The driver cell's emf must exceed the cell being tested, or there is no null point.

Shortcut

Memory clue : "length ∝ potential".

Spot the difference

QuantityFormulaUnit
kV/LV/m
E₁/E₂l₁/l₂—

Practice questions

Easy
  1. 1What is k?
  2. 2Current at the null point?
  3. 3How to compare emfs?
Medium
  1. 4L = 5 m, V = 2.5 V : k?
  2. 5k = 0.2 V/m, l = 4 m : V?
  3. 6l₁ = 80, l₂ = 100, E₂ = 2 V : E₁?
Challenging
  1. 7If k increases, what happens to l?
  2. 8Why is a potentiometer more accurate than a voltmeter?
Show answers (try them yourself first)
  1. 1V/m
  2. 2Zero
  3. 3l₁/l₂
  4. 40.5 V/m
  5. 50.8 V
  6. 6E₁ = 1.6 V
  7. 7l decreases
  8. 8It draws no current

Internal resistance and uses of the potentiometer

From daily life

A phone battery heats up under heavy load because of its internal resistance.

Simple definition

A cell's terminal voltage is V = E − I r.

Potentiometer : internal resistancel₁ (open)l₂ (R)r = R (l₁ − l₂) ÷ l₂

In the figure : Open l₁, with R l₂.

Rule 1: Internal resistance — Cell open : E = k l₁. With R connected, terminal V = k l₂. Then r = R (l₁ − l₂) ÷ l₂.

  • Easy : R = 10 Ω, l₁ = 70, l₂ = 60 ⇒ r = 1.67 Ω.
  • Medium : Always l₁ > l₂.
  • Exam (CET) : Formula for r? (1) R(l₁−l₂)/l₂ (2) R l₁/l₂ (3) R (4) l₁/l₂ → answer 1)

Rule 2: Applications — Comparing emfs, internal resistance, measuring tiny thermocouple emfs, calibrating ammeters.

  • Easy : Thermocouple emfs of µV order need a potentiometer.
  • Medium : Increase sensitivity with a longer wire or smaller k.
  • Exam (CET) : To increase sensitivity? (1) lower k (2) higher k (3) shorter wire (4) larger E → answer 1)

Rule 3: Sensitivity of a potentiometer — A small potential gradient k spreads a small p.d. over a long l, giving accuracy.

  • Easy : 10 m wire, 2 V ⇒ k = 0.2 V/m.
  • Medium : 2 m wire ⇒ k = 1 V/m (less sensitive).
  • Exam (CET) : To reduce k? (1) raise the series resistance (2) raise E (3) shorter wire (4) nothing → answer 1)

Derivation : r = R (l₁ − l₂) ÷ l₂

  • Step 1 : Open : E = k l₁; closed : V = k l₂.
  • Step 2 : E = I(R + r), V = I R ⇒ E ÷ V = (R + r) ÷ R.
  • Step 3 : l₁ ÷ l₂ = (R + r) ÷ R.
  • Step 4 : r = R (l₁ − l₂) ÷ l₂.
Remember

r = R(l₁ − l₂)/l₂; V = E − Ir.

Common mistake / CET trap

Do not interchange l₁ and l₂.

Shortcut

Memory clue : "open longer, closed shorter".

Spot the difference

StateNull pointValue
Openl₁E
With Rl₂V

Practice questions

Easy
  1. 1Formula for r?
  2. 2Which is larger, l₁ or l₂?
  3. 3Benefit of lower k?
Medium
  1. 4R = 5, l₁ = 90, l₂ = 80 : r?
  2. 5R = 20, l₁ = 75, l₂ = 60 : r?
  3. 6L = 8 m, V = 4 V : k?
Challenging
  1. 7R = 12, l₁ = 100, l₂ = 80 : r?
  2. 8E = 1.5 V, l₁ = 75 cm : k?
Show answers (try them yourself first)
  1. 1R(l₁−l₂)/l₂
  2. 2l₁
  3. 3Higher sensitivity
  4. 4r = 0.625 Ω
  5. 5r = 5.0 Ω
  6. 6k = 0.5 V/m
  7. 7r = 3.0 Ω
  8. 8k = 2 V/m

Galvanometer, ammeter and voltmeter

From daily life

A multimeter uses one galvanometer with different resistors attached.

Simple definition

A galvanometer is a sensitive device that detects small currents.

Galvanometer → ammeter and voltmeterGSammeterRvoltmeterS = I_g G ÷ (I − I_g)R = V/I_g − G

In the figure : G + shunt = ammeter; G + large R = voltmeter.

Rule 1: Galvanometer — I_g is the full-scale current; resistance G. Deflection ∝ current. Sensitivity : θ ÷ I = N B A ÷ k.

  • Easy : N = 100, B = 0.1 T, A = 2 cm², k = 10⁻⁶ N m/rad : θ/I = N B A/k = 2 × 10³ rad/A.
  • Medium : More turns ⇒ higher sensitivity.
  • Exam (CET) : To raise sensitivity? (1) raise N, B, A (2) raise k (3) lower N (4) lower B → answer 1)

Rule 2: Ammeter — A small shunt S in parallel with G : S = I_g G ÷ (I − I_g). An ammeter is connected in series; its resistance is ≈ 0.

  • Easy : G = 50 Ω, I_g = 2 mA, I = 1 A ⇒ S = 1.002 × 10⁻¹ Ω.
  • Medium : An ammeter has low resistance.
  • Exam (CET) : Shunt is connected? (1) in parallel (2) in series (3) both (4) none → answer 1)

Rule 3: Voltmeter — A large resistor R in series with G : R = V ÷ I_g − G. A voltmeter is connected in parallel; its resistance is large.

  • Easy : V = 10 V, I_g = 2 mA, G = 50 Ω ⇒ R = 4950 Ω.
  • Medium : An ideal voltmeter has infinite resistance.
  • Exam (CET) : Voltmeter resistance? (1) large (2) small (3) zero (4) equal to G → answer 1)

Derivation : Shunt S = I_g G ÷ (I − I_g)

  • Step 1 : Total I; I_g through G; I − I_g through S.
  • Step 2 : G and S in parallel ⇒ same V : I_g G = (I − I_g) S.
  • Step 3 : Solve for S.
  • Step 4 : S = I_g G ÷ (I − I_g).
Remember

S = I_gG/(I − I_g); R = V/I_g − G.

Common mistake / CET trap

Ammeter in series, voltmeter in parallel — reversing them can damage the meter.

Shortcut

Memory clue : "ammeter — shunt parallel, voltmeter — series R".

Spot the difference

DeviceConnectionResistance
Ammeterseriesvery small
Voltmeterparallelvery large
Shuntparallel to Gsmall

Practice questions

Easy
  1. 1Purpose of a shunt?
  2. 2How is a voltmeter connected?
  3. 3What is I_g?
Medium
  1. 4G = 100, I_g = 1 mA, I = 0.5 A : S?
  2. 5G = 100, I_g = 1 mA, V = 5 V : R?
  3. 6Ammeter resistance, G = 50, S = 0.1 in parallel : R_a?
Challenging
  1. 7S to extend the range 10 times?
  2. 8R to extend a voltmeter range n times?
Show answers (try them yourself first)
  1. 1To extend the current range
  2. 2In parallel
  3. 3Full-scale current
  4. 4S = 0.200 Ω
  5. 5R = 4900 Ω
  6. 6R_a = 0.100 Ω
  7. 7S = G/9
  8. 8R = (n − 1)G in series

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
KCLΣI = 0A
KVLΣE = ΣIRV
Cellsseries nE, nr; parallel E, r/nV, Ω
BridgeP/Q = R/SΩ
Metre bridgeR/S = l/(100−l)Ω
PotentiometerV = kl; E₁/E₂ = l₁/l₂V/m
rR(l₁−l₂)/l₂Ω
ShuntS = I_gG/(I − I_g)Ω
VoltmeterR = V/I_g − GΩ

🎯 Most important points for the exam

  • KCL = charge, KVL = energy conservation.
  • Series nE, nr; parallel E, r/n.
  • Bridge balance P/Q = R/S.
  • Metre bridge R/S = l/(100−l).
  • Potentiometer E₁/E₂ = l₁/l₂.
  • r = R(l₁−l₂)/l₂.
  • Shunt in parallel, voltmeter R in series.
Study mantra

Fix the loop direction → signs → equations → solve → check units.