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NEET Current Electricity: 5 practice MCQs with answers
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Current Electricity is a high-scoring chapter if you can set up circuits quickly. The five MCQs below cover parallel resistors, how resistance changes when a wire is stretched, the resistance of a bulb, a cell with internal resistance and a balanced Wheatstone bridge, all at NCERT Class 12 level.
What this chapter covers in the NEET syllabus
Current Electricity is an NMC Physics unit, matching the Class 12 NCERT chapter of the same name. The topics are:
- Electric current, drift velocity, mobility and Ohm's law.
- Resistivity, conductivity and their dependence on temperature.
- Resistors in series and parallel, and electrical power and energy.
- EMF, internal resistance, and cells in series and parallel.
- Kirchhoff's laws and their use in circuits.
- Wheatstone bridge, metre bridge and the potentiometer.
Five practice MCQs on Current Electricity
Question 1
Resistors of 4 Ω and 12 Ω are connected in parallel. The equivalent resistance is:
- A. 16 Ω
- B. 8 Ω
- C. 3 Ω
- D. 4.5 Ω
Question 2
A wire of resistance R is stretched uniformly to double its length (its volume remains constant). The new resistance is:
- A. 2R
- B. 4R
- C. R/2
- D. R/4
Question 3
A bulb is rated 100 W, 220 V. Its resistance when operating at its rated voltage is:
- A. 22 Ω
- B. 2.2 Ω
- C. 484 Ω
- D. 48.4 Ω
Question 4
A cell of emf 2 V and internal resistance 0.5 Ω is connected to an external resistance of 4.5 Ω. The current in the circuit is:
- A. 0.4 A
- B. 0.5 A
- C. 4 A
- D. 0.44 A
Question 5
In a balanced Wheatstone bridge, P = 2 Ω, Q = 3 Ω and R = 4 Ω, where P and Q are the ratio arms and R is the known resistance. The unknown resistance S is:
- A. 1.5 Ω
- B. 2.67 Ω
- C. 8 Ω
- D. 6 Ω
Answers and explanations
- C. For parallel resistors, the product over the sum gives 4 × 12/(4 + 12) = 48/16 = 3 Ω. A parallel combination is always less than the smallest resistor, so options A, B and D cannot be right.
- B. Resistance is R = ρL/A. With constant volume, doubling L halves A, so R becomes 2 × 2 = 4 times the original. In general, R is proportional to L² at constant volume.
- C. From P = V²/R, R = V²/P = 220²/100 = 48400/100 = 484 Ω. At a lower voltage the bulb would draw less power, but the rated resistance is fixed by the rating.
- A. I = E/(R + r) = 2/(4.5 + 0.5) = 2/5 = 0.4 A. Option D forgets the internal resistance. The terminal voltage is 2 − 0.4 × 0.5 = 1.8 V.
- D. At balance P/Q = R/S, so S = QR/P = 3 × 4/2 = 6 Ω.
How to prepare Current Electricity
- Know the three forms of Ohm's law and power. V = IR, P = VI = I²R = V²/R. Choose the form that uses the quantity that stays fixed, such as current in series or voltage in parallel.
- Simplify circuits in steps. Redraw the circuit, collapse series and parallel groups, and check for a balanced bridge, which removes the middle resistor. Practise ten such networks until the pattern is quick.
- Treat internal resistance as a resistor in series with the emf. Terminal voltage is V = E − Ir when current is drawn, and equals E in an open circuit.
- Apply Kirchhoff's laws carefully. The junction rule is conservation of charge and the loop rule is conservation of energy. Choose a loop direction first and use consistent signs for each element.
- Understand drift velocity. I = n e A vd. Current is the same through all parts of a series conductor, but drift velocity changes if the area changes.
- Learn the instruments. The metre bridge and potentiometer both rely on the balance condition, and the potentiometer draws no current at balance.
This is a chapter where speed matters, so time your practice sets. See the NEET Physics preparation guide for how to fit it into your plan and the NEET syllabus for all units.
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