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Parallel combination of a choke coil and a resistor of resistance $60 Ω$ is connected to an AC source. The currents through the coil, the resistor and the source are $I_{c h} = 3 A, I_{R} = 2.5 A$ and $I_{s} = 4.5 A$ respectively. Therefore,

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Impedance of the coil is $50 Ω$

Impedance of the coil is $150 Ω$

Impedance of the circuit is $\frac{100}{3} Ω$

Heat dissipated in the resistor is 375W

answer is A, C, D.

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Detailed Solution

$4.5 = \sqrt{3^{2} + {2.5}^{2} + 2 \times 2.5 \times 3 c o s θ} = 20.25 = 9 + 6.25 + 15 \cos θ$

$∴ c o s θ = 1 / 3$

$ε_{r m s} = V_{R} = 60 Ω \times 2.5 A = 150 V$

$Z_{c h} = \frac{ε_{r m s}}{I_{c h}} = \frac{150}{3} = 50 Ω, Z_{n e t} = \frac{ε_{r m s}}{I} = \frac{150 V}{4.5} = \frac{100}{3} Ω$

$H_{R} = I_{R}^{2} R = {(2.5)}^{2} \times 60 = 375 W$ $H_{c h} = ε_{r m s} I_{c h} c o s θ = 150 \times 3 \times \frac{1}{3} = 150 W$

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