Current Electricity Solved Examples (Class 12 Physics)
Analyze circuits using Ohm's law, resistivity, and EMF-internal resistance concepts. These problems develop competence in complex circuit analysis and elec
TL;DR: Analyze circuits using Ohm's law, resistivity, and EMF-internal resistance concepts. These problems develop competence in complex circuit analysis and…
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Analyze circuits using Ohm's law, resistivity, and EMF-internal resistance concepts. These problems develop competence in complex circuit analysis and elec
Current Electricity — Solved Numerical Examples (Step by Step)
Example 1: A wire has a length of 5 m and cross-sectional area of 2 × 10^-6 m². If its resistivity is 1.7 × 10^-8 Ω·m, find its resistance.
Solution: Using R = ρL/A
R = (1.7 × 10^-8 × 5) / (2 × 10^-6)
R = (8.5 × 10^-8) / (2 × 10^-6)
R = 4.25 × 10^-2 Ω = 0.0425 Ω
Example 2: A battery with EMF 12 V and internal resistance 2 Ω is connected to an external resistance of 10 Ω. Find the current in the circuit.
Solution: Total resistance = External resistance + Internal resistance = 10 + 2 = 12 Ω
Using I = E/(R + r)
I = 12 / 12 = 1 A
Example 3: For the circuit in the previous problem, find the terminal voltage across the external resistance.
Solution: Current I = 1 A
Terminal voltage V = E - Ir = 12 - 1 × 2 = 10 V
Alternatively, V = I × R_external = 1 × 10 = 10 V
Example 4: A heater with resistance 20 Ω is connected to 220 V mains. Find the power consumed and current drawn.
Solution: Using Ohm's law: I = V/R = 220 / 20 = 11 A
Power P = VI = 220 × 11 = 2420 W
Alternatively, P = V²/R = 220² / 20 = 48400 / 20 = 2420 W
Example 5: Three cells, each with EMF 2 V and internal resistance 1 Ω, are connected in series. They are connected to an external resistance of 15 Ω. Find the current.
Solution: Total EMF = 3 × 2 = 6 V
Total internal resistance = 3 × 1 = 3 Ω
Total resistance = 15 + 3 = 18 Ω
Current I = Total EMF / Total resistance = 6 / 18 = 1/3 A ≈ 0.33 A
Example 6: A copper wire of length 100 m and cross-sectional area 10^-6 m² has resistivity 1.6 × 10^-8 Ω·m. If current of 2 A flows through it, find the potential difference across the wire.
Solution: Resistance R = ρL/A = (1.6 × 10^-8 × 100) / (10^-6)
R = (1.6 × 10^-6) / (10^-6) = 1.6 Ω
Potential difference V = IR = 2 × 1.6 = 3.2 V
Tips
- Resistivity ρ is a material property independent of shape; R = ρL/A shows how geometry affects resistance.
- For batteries: E is EMF (open circuit), V = E - Ir is terminal voltage (closed circuit), where r is internal resistance.
- Power in resistor can be calculated three ways: P = VI = I²R = V²/R; choose based on known quantities.
Frequently Asked Questions
Why does a battery have internal resistance?
Internal resistance arises from the material of the battery and opposes current flow. It causes the terminal voltage to drop below EMF when current flows.
What is the difference between EMF and terminal voltage?
EMF (ε) is the total energy per unit charge provided by the battery. Terminal voltage (V) is the potential difference available at the battery terminals, reduced by the voltage drop across internal resistance.
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