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(A) Given: Resistance $R = 90 \Omega$, Supply voltage $V = 120 	ext{ V}$, Frequency $f = 60 	ext{ Hz}$, Voltage across resistor $V_R = 36 	ext{ V}$

Vedclass · Accepted Answer

$0.76$

Vedclass · Answer

(A) Given: Resistance $R = 90 \Omega$, Supply voltage $V = 120 	ext{ V}$, Frequency $f = 60 	ext{ Hz}$, Voltage across resistor $V_R = 36 	ext{ V}$.The current in the series circuit is $I = \frac{V_R}{R} = \frac{36}{90} = 0.4 	ext{ A}$.The impedance of the circuit is $Z = \frac{V}{I} = \frac{120}{0.4} = 300 \Omega$.We know $Z = \sqrt{R^2 + X_L^2}$, so $300 = \sqrt{90^2 + X_L^2}$.Squaring both sides: $90000 = 8100 + X_L^2$.$X_L^2 = 90000 - 8100 = 81900$.$X_L = \sqrt{81900} \approx 286.18 \Omega$.Since $X_L = 2 \pi f L$, we have $L = \frac{X_L}{2 \pi f} = \frac{286.18}{2 	imes 3.14 	imes 60} = \frac{286.18}{376.8} \approx 0.76 	ext{ H}$.

$A$ coil of negligible resistance is connected in series with a $90 \Omega$ resistor across a $120 \text{ V}, 60 \text{ Hz}$ supply. $A$ voltmeter reads $36 \text{ V}$ across the resistor. The inductance of the coil is: (in $\text{ H}$)

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