A current of 4 A flows in a coil when connected to a 12 V d.c. source. If the same coil is connected to a 12 V, 50 rad/s, a.c. source, a current of 2.4 A flows in the circuit. Determine the inductance of the coil. Also find the power developed in the circuit if a 2500 μF condenser is connected in series with the coil.

Solution

Current flowing in the coil, I = 4 A
Rms voltage = 12 V
Angular frequency of the AC source = 50 rad/sec
Capacitance, C = 2500 μF = 2500

\times

10^-6 F

When the coil is connected to a d.c. source, its resistance R is given by

R = \frac{V}{I} = \frac{12}{4} = 3 Ω

When it is connected to a.c. source, the impedance Z of the coil is given by

Z = \frac{V_{rms}}{I_{rms}} = \frac{12}{2.4} = 5 Ω

For a coil,

Z = \sqrt{[R^{2} + (ωL)^{2}]}

∴

5 = \sqrt{[(3)^{2} + (50 L)^{2}]}

25 = [(3)^{2} + (50 L)^{2}]

Solving we get,
Inductance, L = 0.08 henry.

When the coil is connected with a condenser in series, the impedance Z' is given by

Z' = \sqrt{[R^{2} + {(ωL - \frac{1}{ωC})}^{2}]}

= {[(3)^{2} + {(50 \times 0.08 - \frac{1}{50 - 2500 \times 10^{- 6}})}^{2}]}^{1 / 2} = 5 ohm

Power developed in the circuit,

P = V_{rms} \times I_{rms} \times \cos θ

where

\cos ϕ = R / Z' = 3 / 5 = 0.6

Therefore,
P = 12

\times 2.4 \times 0.6

17.28 watt

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