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 wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:
- A. R/64
- B. R/32
- C. R/16
- D. R/8
Answer: C. R/16
Resistance of each piece = R/8. Parallel combination of 4 pieces: 1/Rp = 1/(R/8) + 1/(R/8) + 1/(R/8) + 1/(R/8) = 32/R. So Rp = R/32. Two such sets in series: R_net = R/32 + R/32 = 2R/32 = R/16.
Resistance of each piece = R/8. Parallel combination of 4 pieces: 1/Rp = 1/(R/8) + 1/(R/8) + 1/(R/8) + 1/(R/8) = 32/R. So Rp = R/32. Two such sets in series: R_net = R/32 + R/32 = 2R/32 = R/16.
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is:
- A. 1.5 A
- B. 2.0 A
- C. 2.5 A
- D. 3.0 A
Answer: B. 2.0 A
Based on the official key provided for question 36.
Based on the official key provided for question 36.
A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf’s is :
- A. 5 : 4
- B. 3 : 4
- C. 3 : 2
- D. 5 : 1
Answer: C. 3 : 2
E1+E2 is proportional to 50, E1-E2 is proportional to 10. Solving the ratio E1/E2 gives 60/40 = 3/2.
E1+E2 is proportional to 50, E1-E2 is proportional to 10. Solving the ratio E1/E2 gives 60/40 = 3/2.
The charge flowing through a resistance R varies with time t as Q = at – bt², where a and b are positive constants, The total heat produced in R is:
- A. a³/b³R
- B. a³/b²R
- C. a³/bR
- D. a³R/6b
Answer: D. a³R/6b
Current I = dQ/dt = a - 2bt. Heat is produced until I = 0, which occurs at t = a/(2b). Total heat = integral of I²R dt from 0 to a/(2b), resulting in a³R/(6b).
Current I = dQ/dt = a - 2bt. Heat is produced until I = 0, which occurs at t = a/(2b). Total heat = integral of I²R dt from 0 to a/(2b), resulting in a³R/(6b).
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.
Which of the following graph represents the variation of resistivity (ρ) with temperature (T) for copper?
- A. Line with positive slope starting from near origin
- B. Hyperbola
- C. Exponential
- D. Constant
Answer: A. Line with positive slope starting from near origin
Copper is a metal; resistivity increases linearly with temperature.
Copper is a metal; resistivity increases linearly with temperature.