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Q.

A coil is connected to an alternating emf of voltage $24 V$ and of frequency $50 H z .$ The reading on the ammeter connected to the coil in series is $10 m A .$ If a $1 μ F$ capacitor is connected to the coil in series the ammeter shows $10 m A$ again. What would be the reading on a dc ammeter (in A) if the coil was connected to a $180 V$ dc voltage supply? $(Take π^{2} = 10)$

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Detailed Solution

$\begin{array}{l} \frac{24}{10 \times 10^{- 3}} = z = \sqrt{(ω L)^{2} + R^{2}} \\ \sqrt{R^{2} + (ω L)^{2}} = \sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}} \\ (ω L) = - ω L + \frac{1}{ω C} \\ L = \frac{1}{2 ω^{2} C} V = \frac{1}{2 \times 100 π \times 100 π \times 10^{- 6}} = 5 H \\ (2400)^{2} = (500 π)^{2} + R^{2} \\ R = \sqrt{(2400)^{2} - (5 π \times 100)^{2}} = 10 \sqrt{(24)^{2} - 25 π^{2}} \\ = 10 \times \sqrt{326} ≅ 180 \\ I = 1 A \end{array}$

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A coil is connected to an alternating emf of voltage 24 V and of frequency 50 Hz. The reading on the ammeter connected to the coil in series is 10 mA. If a 1μF capacitor is connected to the coil in series the ammeter shows 10 mA again. What would be the reading on a dc ammeter (in A) if the coil was connected to a 180 V dc voltage supply? (Take π2=10)