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A parallel combination of an inductor coil and a resistance of 60 ohm is connected to an ac source. The current in the coil, current in the resistance and the source current are respectively 3A, 2.5 A and 4.5 A. Therefore,

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Impedance of the circuit is 33.3 ohm

Kirchhoff’s current law is NOT applicable to ac circuits

Impedance of the coil is 50 ohm

Electric power dissipated in the coil is 150 watt

answer is B, D.

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Detailed Solution

$V_{R_{r m s}} \Rightarrow 60 \times \frac{2.5}{10} = 150 v o l t$

$V_{r m s} = V_{R_{r m s}} = 150 V$

$Z = \frac{V_{r m s}}{i_{r m s}} \Rightarrow \frac{150}{4.5} \Rightarrow \frac{150}{4.5} \times 10 = \frac{100}{3} = 33.3 Ω ∴$ Ans(d)

$V_{L_{r m s}} = 150 = i_{r m s} \times X_{L}$

$X_{L} = \frac{150}{3} = 50 Ω ∴$ Ans.(b)

So correct options are (b) & (d)

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