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Resistors, capacitors and inductors are connected to either a variable DC source or a 50 Hz AC source as shown. A current I (steady-state for DC, rms for AC) flows, producing voltages V1 and V2 across the two indicated elements. Which circuit satisfies the relation 'I is not zero AND V2 > V1', given the elements 6 mH, 3 microF, 2 ohm and source 50 Hz where applicable?

1. (P) 6mH and 3microF in series with variable DC source, V1 across 6mH and V2 across 3microF
2. (Q) 6mH and 2ohm in series with variable DC source, V1 across 6mH and V2 across 2ohm
3. (R) 6mH and 2ohm in series with AC source, V1 across 6mH and V2 across 2ohm
4. (S) 6mH and 3microF in series with AC source, V1 across 6mH and V2 across 3microF

Correct answer: (S) 6mH and 3microF in series with AC source, V1 across 6mH and V2 across 3microF

Solution

Condition required: current must flow (I != 0) and V2 must exceed V1. (P) is L-C in DC: a capacitor blocks DC, so steady-state I = 0; rejected. (Q) is L-R in DC: I != 0 but the inductor has V1 = 0 in steady DC and V2 = IR, so V2 > V1 but here V1 = 0 makes this a 'V1 = 0, V2 = V' case more than a strict V2 > V1 comparison of two nonzero drops. (R) is L-R in AC: V1 = I*omega*L = I*(2*pi*50*6e-3) = I*1.885 ohm, V2 = I*2 ohm; here V2 (2 ohm) > V1 (1.885 ohm) is only marginal. (S) is L-C in AC: V1 = I*X_L = I*omega*L = I*1.885 ohm, V2 = I*X_C = I/(omega*C) = I/(2*pi*50*3e-6) = I*1061 ohm. Since X_C >> X_L, V2 >> V1 with I != 0, cleanly satisfying I != 0 and V2 > V1. So (S) is the circuit that best matches the relation.

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