Free NEET Physics practice
Current Electricity: MCQs with Solutions
Start by deciding which circuit elements share current and which share potential difference. These original problems use ideal wires and steady direct current unless stated otherwise. Write the equivalent resistance before calculating current or power.
5 original practice questions — not previous-year paper reproductions. No account required.
By NEET Ninja · Updated
Before you answer
- Series resistors carry the same current and their resistances add. Parallel resistors have the same voltage and their conductances add.
- For a uniform conductor, R = ρL/A. Resistivity is a material property at a specified temperature; resistance also depends on geometry.
- Distinguish a cell’s emf from its terminal voltage under load. With internal resistance r, current is E/(R + r) and terminal voltage is E − Ir.
Questions and worked answers
Choose an option before reading the answer. For a wrong answer, identify the step or concept that changed the result.
QUESTION 1
A 4 Ω resistor and an 8 Ω resistor are connected in series across an ideal 24 V supply. What current flows?
- 1 A
- 2 A
- 3 A
- 6 A
Answer: B — 2 A
The series equivalent resistance is 4 + 8 = 12 Ω. Applying Ohm’s law to the whole circuit gives I = 24/12 = 2 A. This same current passes through both resistors. Their individual voltage drops are 8 V and 16 V, which add to 24 V.
QUESTION 2
What is the equivalent resistance of 6 Ω and 3 Ω resistors connected in parallel?
- 2 Ω
- 3 Ω
- 4.5 Ω
- 9 Ω
Answer: A — 2 Ω
For two resistors in parallel, 1/R = 1/6 + 1/3 = 1/2, so R = 2 Ω. The result must be less than the smaller branch resistance, 3 Ω. Adding the resistances to obtain 9 Ω would apply to a series arrangement, not a parallel one.
QUESTION 3
A uniform wire is stretched to twice its original length without changing its volume or resistivity. How does its resistance change?
- It becomes R/2
- It remains R
- It becomes 2R
- It becomes 4R
Answer: D — It becomes 4R
Constant volume means AL stays fixed. Doubling the length halves the cross-sectional area. Substituting into R = ρL/A gives R′ = ρ(2L)/(A/2) = 4R. Counting only the length increase misses the simultaneous decrease in cross-sectional area.
QUESTION 4
An ideal 12 V supply maintains 12 V across a 6 Ω resistor. What power is dissipated?
- 2 W
- 12 W
- 24 W
- 72 W
Answer: C — 24 W
Use P = V²/R because the voltage across the resistor is specified. Thus P = 12²/6 = 24 W. Alternatively I = V/R = 2 A and P = VI = 24 W. The value 2 is the current in amperes, not the power in watts.
QUESTION 5
A cell of emf 6 V and internal resistance 1 Ω supplies an external resistance of 5 Ω. What is its terminal voltage?
- 1 V
- 5 V
- 6 V
- 7 V
Answer: B — 5 V
The total circuit resistance is 5 + 1 = 6 Ω, so the current is 6/6 = 1 A. The terminal voltage is the voltage across the external resistor: V = 1 × 5 = 5 V. The remaining 1 V is the internal drop Ir, so terminal voltage is less than emf during discharge.
Study source and corrections
Use the matching chapter in NCERT Class XII Physics, Part I — Current Electricity for the underlying concepts. These questions are written for NEET Ninja, not endorsed or supplied by NCERT or NTA.
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