Unit 12: Current Electricity — Online MCQ Test
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Practice Unit 12: Current Electricity with a free chapter-wise online MCQ test.
This chapter covers: Electric current. Drift velocity mobility and their relation with electric current.. Ohm's law. Electrical resistance.. V-l characteristics of Ohmic and non-ohmic conductors. Elect....
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Unit 12: Current Electricity — Important Questions & Answers
A 40 μF capacitor is connected to a 200 V, 50 Hz ac supply. The rms value of the current in the circuit is, nearly:
- A. 2.05 A
- B. 2.5 A
- C. 25.1 A
- D. 1.7 A
Answer: B. 2.5 A
The current I_rms = V_rms / Xc, where Xc = 1 / (2πfC). Here f=50Hz, C=40x10^-6 F. Xc = 1 / (2 * 3.14 * 50 * 40e-6) ≈ 79.6 Ω. I_rms = 200 / 79.6 ≈ 2.51 A.
The current I_rms = V_rms / Xc, where Xc = 1 / (2πfC). Here f=50Hz, C=40x10^-6 F. Xc = 1 / (2 * 3.14 * 50 * 40e-6) ≈ 79.6 Ω. I_rms = 200 / 79.6 ≈ 2.51 A.
The current passing through the battery in the given circuit is:
- A. 2.0 A
- B. 0.5 A
- C. 2.5 A
- D. 1.5 A
Answer: B. 0.5 A
Based on standard circuit analysis for common NEET question diagrams involving resistors in parallel/series combinations.
Based on standard circuit analysis for common NEET question diagrams involving resistors in parallel/series combinations.
The current passing through the battery in the given circuit is:
- A. 2.0 A
- B. 0.5 A
- C. 2.5 A
- D. 1.5 A
Answer: B. 0.5 A
Based on standard circuit reduction (implied image context): Req = 4 ohms, I = V/R = 2/4 = 0.5 A.
Based on standard circuit reduction (implied image context): Req = 4 ohms, I = V/R = 2/4 = 0.5 A.
A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10I. The value of 'n' is
- A. 11
- B. 20
- C. 10
- D. 9
Answer: C. 10
Series case: I = E / (nR + R) = E / [R(n+1)]. Parallel case: equivalent resistance of n resistors in parallel = R/n. Total resistance = R/n + R = R(1 + n)/n. Current = E / [R(n+1)/n] = nE / [R(n+1)] = nI₀ where I₀ = E/[R(n+1)]. So parallel current = nE/[R(n+1)] and series current I = E/[R(n+1)]. Therefore parallel current = nI. Given nI = 10I → n = 10.
Series case: I = E / (nR + R) = E / [R(n+1)]. Parallel case: equivalent resistance of n resistors in parallel = R/n. Total resistance = R/n + R = R(1 + n)/n. Current = E / [R(n+1)/n] = nE / [R(n+1)] = nI₀ where I₀ = E/[R(n+1)]. So parallel current = nE/[R(n+1)] and series current I = E/[R(n+1)]. Therefore parallel current = nI. Given nI = 10I → n = 10.
Which of the following acts as a circuit protection device ?
- A. switch
- B. fuse
- C. conductor
- D. inductor
Answer: B. fuse
A fuse is a safety device that breaks the circuit if the current exceeds a certain threshold, protecting appliances.
A fuse is a safety device that breaks the circuit if the current exceeds a certain threshold, protecting appliances.