Electricity – Numerical Problems with Stepwise Solutions

CBSE Class 10 Physics — Chapter 11: Electricity — 20 Numerical Problems with Solutions

Content Bank — Important Formulas (Chapter: Electricity)

Ohm's Law:
V = I·R

Resistance:
R = V/I

Resistivity relation:
R = ρ·L/A

Series resistors:
R_eq = R1 + R2 + ...

Parallel resistors:
1/R_eq = 1/R1 + 1/R2 + ...

Power:
P = V·I = I²R = V²/R

Energy:
E = P·t (kWh = kW × h)

Cell with internal r:
V_terminal = E − I·r

Heating (Joule):
H = I²·R·t

Max power transfer:
Max when R_load = r_internal

Tip: Always write units (Ω, V, A, W, J, s) and show each algebraic step — CBSE awards method marks.

Numerical Problems with Stepwise Solutions (20) — Topic-wise

Q1 (Basics):

Problem: A conductor has a potential difference of 5.0 V across its ends and a current of 0.25 A flows through it. Find its resistance.

Use R = V / I.
R = 5.0 V / 0.25 A = 20 Ω.
Answer: 20 Ω.

Q2 (Ohm's law):

Problem: A wire obeying Ohm's law carries 2.0 A when 6 V is applied. What current will flow when 15 V is applied to the same wire?

Find resistance: R = V/I = 6 / 2 = 3 Ω.
For 15 V, I = V / R = 15 / 3 = 5 A.
Answer: 5.0 A.

Q3 (Graphical check):

Problem: A conductor shows I = 0.5 A at V = 2 V and I = 1.5 A at V = 6 V. Is it ohmic? Also find R.

Compute R for each point: R1 = 2/0.5 = 4 Ω; R2 = 6/1.5 = 4 Ω.
Since R is constant, conductor is ohmic.
Answer: Yes, ohmic; R = 4 Ω.

Q4 (Charge-current-time):

Problem: A current of 3 A flows for 10 s. How much charge passes through the conductor?

Use Q = I × t = 3 A × 10 s = 30 C.
Answer: 30 coulombs.

Q5 (Series):

Problem: Two resistors 8 Ω and 12 Ω are connected in series across 20 V. Find the current and voltage across each resistor.

R_total = 8 + 12 = 20 Ω.
Current I = V / R_total = 20 / 20 = 1 A.
Voltage across 8 Ω: V1 = I × 8 = 1 × 8 = 8 V. Across 12 Ω: V2 = 12 V.
Answer: I = 1 A; V1 = 8 V, V2 = 12 V.

Q6 (Parallel):

Problem: Two resistors 4 Ω and 6 Ω are connected in parallel across 12 V. Find the currents through each and the total current.

Voltage across each = 12 V.
I1 = 12 / 4 = 3 A; I2 = 12 / 6 = 2 A.
Total I = 3 + 2 = 5 A.
Answer: I1 = 3 A, I2 = 2 A, I_total = 5 A.

Q7 (Parallel equivalent):

Problem: Find equivalent resistance of 3 resistors 6 Ω, 6 Ω and 3 Ω in parallel.

Compute reciprocals: 1/R = 1/6 + 1/6 + 1/3 = (1+1+2)/6 = 4/6 = 2/3.
So R_eq = 3/2 = 1.5 Ω.
Answer: 1.5 Ω.

Q8 (Series-parallel mix):

Problem: R1 = 6 Ω in series with parallel R2 = 3 Ω and R3 = 6 Ω. Find total current from a 12 V battery.

R2||R3 = 1 / (1/3 + 1/6) = 2 Ω.
R_total = 6 + 2 = 8 Ω. I_total = 12 / 8 = 1.5 A.
Answer: 1.5 A.

Q9 (Internal resistance basic):

Problem: A cell of emf 12 V with internal resistance r supplies 2 A to a 5 Ω resistor. Find r.

Terminal voltage across load V = I × R = 2 × 5 = 10 V.
Emf E = V + I·r → r = (E − V)/I = (12 − 10)/2 = 1 Ω.
Answer: 1.0 Ω.

Q10 (Terminal voltage):

Problem: A 9 V battery with internal resistance 0.5 Ω supplies current 2 A. What is terminal voltage?

V_term = E − I·r = 9 − 2×0.5 = 9 − 1 = 8 V.
Answer: 8.0 V.

Q11 (Max power idea — calculation):

Problem: A cell has E = 12 V and internal resistance r = 2 Ω. What load gives maximum power and what is that power?

Max power when R_load = r = 2 Ω.
Current at R=2: I = E/(R+r) = 12/(2+2) = 12/4 = 3 A.
Power in load P = I²R = 3² × 2 = 9 × 2 = 18 W.
Answer: R_load = 2 Ω; P_max = 18 W.

Q12 (Power calculation):

Problem: A heater of resistance 20 Ω connected to 220 V. Find current and power.

I = V/R = 220 / 20 = 11 A.
P = V × I = 220 × 11 = 2420 W ≈ 2.42 kW.
Answer: I = 11 A; P = 2420 W.

Q13 (Energy & cost):

Problem: A 1.5 kW heater runs for 2 h. Calculate energy used (kWh) and cost if tariff is ₹8.50/kWh.

Energy = 1.5 kW × 2 h = 3.0 kWh.
Cost = 3.0 × 8.50 = ₹25.50.
Answer: 3.0 kWh; cost ₹25.50.

Q14 (Heating for time):

Problem: A toaster draws 5 A at 230 V. How much heat is produced in 10 minutes? (in kJ)

Power P = V×I = 230 × 5 = 1150 W = 1150 J/s.
Time t = 10 min = 600 s. Energy = P×t = 1150 × 600 = 690000 J = 690 kJ.
Answer: 690 kJ.

Q15 (Resistivity relation):

Problem: A wire of length 2 m and cross-sectional area 1.0 × 10^-6 m² has resistivity 1.6 × 10^-8 Ω·m. Find its resistance.

Use R = ρL/A = (1.6×10^-8 × 2) / (1.0×10^-6) = (3.2×10^-8)/(1.0×10^-6) = 3.2×10^-2 Ω.
Answer: 0.032 Ω (32 mΩ).

Q16 (Length scaling):

Problem: Wire A has resistance 1 Ω. Wire B is made from same material and area but twice the length. Find resistance of B.

R ∝ L, so R_B = 2 × R_A = 2 × 1 Ω = 2 Ω.
Answer: 2 Ω.

Q17 (Area scaling):

Problem: A wire length L has resistance R. If its cross-sectional area is doubled (same length/material), what is new resistance?

R = ρL/A → doubling A halves R. So R_new = R/2.
Answer: R/2.

Q18 (Combined problem):

Problem: A 12 V battery is connected to two resistors 3 Ω and 6 Ω in parallel. Find total current drawn and power dissipated in each resistor.

Voltage across each = 12 V. Currents: I3 = 12/3 = 4 A; I6 = 12/6 = 2 A.
Total current = 4 + 2 = 6 A.
Power: P3 = V×I3 = 12×4 = 48 W; P6 = 12×2 = 24 W.
Answer: I_total = 6 A; P3 = 48 W, P6 = 24 W.

Q19 (Short-circuit estimation):

Problem: A battery of emf 9 V and internal resistance 0.1 Ω is shorted (external R≈0). Estimate the short-circuit current.

Short-circuit current I ≈ E / r = 9 / 0.1 = 90 A.
Answer: ≈ 90 A (dangerously large).

Q20 (Design & divider):

Problem: Design a voltage divider from 12 V to get 5 V using two resistors R1 (top) and R2 (bottom) in series. Choose standard values with R_total ≈ 10 kΩ.

We need V_out = 5 V = 12 × R2/(R1+R2) → R2/(R1+R2) = 5/12 ≈ 0.4167.
So ratio R1:R2 = (1 − 0.4167):0.4167 = 0.5833:0.4167 ≈ 1.4 : 1 → R1 ≈ 1.4 R2.
If R_total = 10 kΩ, choose R2 = 10k × 0.4167 ≈ 4167 Ω ≈ use standard 4.3 kΩ (or 4.7 kΩ with small error). Then R1 ≈ 10k − 4.3k = 5.7 kΩ (use 5.6 kΩ standard).
Check V_out ≈ 12 × 4.3/(5.7+4.3)=12×4.3/10≈5.16 V (close). Using 4.7 kΩ & 5.3 kΩ produces slightly different voltage; pick pair 5.6 kΩ & 4.3 kΩ for V_out≈5.16 V.
Answer: Choose R1 ≈ 5.6 kΩ and R2 ≈ 4.3 kΩ (R_total ≈ 9.9 kΩ) giving V_out ≈ 5.16 V (close to 5 V).

Electricity – Numerical Problems with Stepwise Solutions

Content Bank — Important Formulas (Chapter: Electricity)

Numerical Problems with Stepwise Solutions (20) — Topic-wise

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