Question

In an ac circuit, values of voltage and current change every instant. Therefore, power of an ac circuit at any instant is the product of instantaneous voltage (E) and instantaneous current (I). The average power supplied to a pure resistance R over a complete cycle of ac is P = E_rms. I_rms. When the circuit is inductive, average power/ cycle = E_rms I_rmscosφ,

Where φ is the phase angle between alternating voltage and alternating current in the circuit. In an ac circuit, 800 mH inductor and a 60µf capacitor are connected in series with 15ohm resistance. The ac supply to the circuit is 230V, 50 Hz.

(i).The average power transferred per cycle to resistance is : 

(a). 2.05 W 

(b). 20.05 W 

(c). zero 

(d). 230 W

(ii).The average power transferred per cycle to inductor is : 

(a). Zero 

(b). 40.10 W 

(c) 20.05 W 

(d). 2.05 W 

(iii).The average power transformed per cycle to capacitor is : 

(a). 230W 

(b). 220 W 

(c) 40 W 

(d) zero 

(iv).The total power absorbed per cycle by all the three-circuit elements is : 

(a). 230W 

(b). 690W 

(c) 60.15W 

(d)20.05W 

(v).The electrical energy spent in running the circuit for 1 h is : 

(a). 20.05 J 

(b) 20.05 kwh 

(c) 7.22× 10⁴J 

(d) zero

Answer 1

Here,

L= 800mH 

= 800 × 10⁻³H 

= 0.8H 

C = 60 µf 

= 60 × 10⁻⁶ F 

R = 15Ω 

E_rms = 230V 

v = 50Hz 

X_L = ω_L 

= 2πvL 

= 2 × 3.14 × 50 × 0.8 

= 251.43Ω

X_c = \(\frac{1}{ωC}\)

= \(\frac{1}{2\pi v c}\) 

= \(\frac{1}{2\times 3.14\times 50\times 60\times 10^{-6}}\) 

= 53.03Ω

(i). Option : (b)

I_rms = \(\frac{E_{rms}}{Z}\) 

= \(\frac{230}{198.97}\) 

= 1.156A

∴ Average power transferred/cycle to resistance :

P_R = \(I^2_{Rms}\) × R 

= (1.156)² × 15 

= 20.05 Watt

(ii). Option : (a)

Average power transferred/cycle to inductor :

P_L = E_rms.I_rms cos φ 

In an inductor, 

Phase difference φ = 90° 

∴P_L= E_rms .I_rms. cos 90° 

= 0 

(iii). Option : (d) 

Average power transferred/cycle to capacitor is : 

P_C = E_rms.I_rms.cos 90° 

= 0 

[since, in a capacitor, phase difference, φ = 90°]

(iv). Option : (d)

Total power absorbed/ cycle, 

P = P_R + P_L + P_C 

= 20.05 + 0 + 0 

= 20.05 Watt.

(v). Option : (c) 

As energy spent = Power × time 

= 20.05 × 60 × 60 

= 7.22 × 10⁴ Joule.