A coil having a resistance of 5 $Ω$ and an inductance of 0.02 H is arranged in parallel with another coil having a resistance of 1 $Ω$ and an inductance of 0.08 H. Calculate the power absorbed (in watt) when a voltage of 100 V at 50 Hz is applied.

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Detailed Solution

$\begin{array}{l} X_{L_{1}} = {ωL}_{1} = (2 π \times 50) (0 .02) = 6 .28 Ω \\ Z_{1} = \sqrt{R_{1}^{2} + X_{L_{1}}^{2}} \\ = \sqrt{(5)^{2} + (6 .28)^{2}} = 8 .0 Ω \end{array} \begin{array}{l} P_{1} = {(I_{rms})}_{1} V_{rms} \cos ϕ = (\frac{100}{8}) (100) (\frac{5}{8}) \\ = 781 .25 W \\ X_{L_{2}} = {ωL}_{2} = (2 π \times 50) (0 .08) = 25 .13 Ω \\ Z_{2} = \sqrt{R_{2}^{2} + X_{L_{2}}^{2}} = 25 .15 Ω \\ P_{2} = {(i_{r m s})}_{2} {(V_{r m s})}_{2} \cos ϕ_{2} \\ = (\frac{100}{25 .15}) (100) (\frac{1}{25 .15}) = 15.82 W \\ P_{total} = P_{1} + P_{2} = 797.07 W \end{array}$