A two-element series circuit is connected across an a.c. source e = 20

1.	A two-element series circuit is connected across an a.c. source e = 200√2 sin (ωt + 20°) V. The current in the circuit then is found to be i = 10 √2 cos (314 t − 25°) A. Determine the parameters of the circuit.
Answer» The current can be written as i = 10√2 sin (314 t − 25° + 90°) = 10√2 sin (314 t + 65°). It is seen that applied voltage leads by 20° and current leads by 65° with regards to the reference quantity, their mutual phase difference is = 65° −(20°) = 45°. Hence, p.f. = cos 45° = 1/ √2 (lead). Now, V_m = 200√2 and I_m = 10√2 ∴ Z = V_m/I_m = 200 √2/10√2 = 20 Ω R = Z cos φ = 20/√2Ω = 14.1 Ω ; X_c = Z sin φ = 20√2 = 14.1 Ω Now, f = 314/2π = 50 Hz. Also, X_c = 1/2π fC ∴ C = 1/2π × 50 × 14.1 = 226μF Hence, the given circuit is an R-C circuit.

A two-element series circuit is connected across an a.c. source e = 200√2 sin (ωt + 20°) V. The current in the circuit then is found to be i = 10 √2 cos (314 t − 25°) A. Determine the parameters of the circuit.

Answer»

The current can be written as i = 10√2 sin (314 t − 25° + 90°) = 10√2 sin (314 t + 65°).

It is seen that applied voltage leads by 20° and current leads by 65° with regards to the reference quantity, their mutual phase difference is = 65° −(20°) = 45°.

Hence, p.f. = cos 45° = 1/ √2 (lead).

Now, V_m = 200√2 and I_m = 10√2

∴ Z = V_m/I_m = 200 √2/10√2 = 20 Ω

R = Z cos φ = 20/√2Ω = 14.1 Ω ; X_c = Z sin φ = 20√2 = 14.1 Ω

Now, f = 314/2π = 50 Hz. Also, X_c = 1/2π fC

∴ C = 1/2π × 50 × 14.1 = 226μF

Hence, the given circuit is an R-C circuit.

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