A 12 ohm resistance and an inductance of 0.05/ $π$ henry with negligible resistance are connected in series. Across the end of this circuit is connected a 130 volt alternating voltage of frequency 50 cycles/second. Calculate the alternating current in the circuit and potential difference across the resistance and that across the inductance.

Solution

Given,
Resistance, R = 12

Ω

Inductance, L = 0.05/

π

H
Rms voltage, V_rms = 130 V
Angular frequency,

ω

= 50 cycles/sec

The impedance of the circuit is given by

Z = \sqrt{(R^{2} + ω^{2} L^{2})} = \sqrt{[R^{2} + (2 πfL)^{2}]} = \sqrt{[(12)^{2} + {\{2 \times 3.14 \times 50 \times (0.05 / 3.14)\}}^{2}]} = \sqrt{(144 + 25)} = 13 ohm

Current in the circuit,

i = \frac{E}{Z} = \frac{130}{13} = 10 amp .

Potential difference across resistance,

V_R = iR
= 10 × 12
= 120 volt

Inductive reactance of coil, X_L = ωL =

2 πfL

∴

X_{L} = 2 π \times 50 \times (\frac{0.05}{π}) = 5 ohm .

Potential difference across inductance

V_L = i × X_L
= 10 × 5
= 50 volt.

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