An alternating voltage $V = 6 sin (500 t)$ is applied across the series combination of a $0.2$ henry inductance and a resistance of $100 o h m$ . Calculate:
(i) phase difference between the current and the applied voltage
(ii) average power consumed in the circuit.

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Solution

Supply voltage is given as $V = 6 sin (500 t)$
Comparing it with $V = V_{o} sin w t$
We get, $w = 500$ $s^{- 1}$
Also, peak voltage $V_{o} = 6$ Volts
Inductive reactance $X_{L} = w L = 500 \times 0.2 = 100 Ω$
Impedance of circuit $Z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{100^{2} + 100^{2}} = 100 \sqrt{2} Ω$
(i) Power factor $cos ϕ = \frac{R}{Z} = \frac{100}{100 \sqrt{2}} = \frac{1}{\sqrt{2}}$
$⟹ ϕ = 45^{o}$
(ii) So, rms value of voltage $V_{r m s} = \frac{V_{o}}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3 \sqrt{2}$ Volts
Average power consumed $P_{a v g} = \frac{V_{r m s}^{2}}{Z} cos ϕ$
$⟹ P_{a v g} = \frac{(3 \sqrt{2})^{2}}{100 \sqrt{2}} \times \frac{1}{\sqrt{2}} = 0.09 W$

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An alternating voltage V=6sin(500 t) is applied across the series combination of a 0.2 henry inductance and a resistance of 100 ohm. Calculate:(i) phase difference between the current and the applied voltage(ii) average power consumed in the circuit.

An alternating voltage $V = 6 sin (500 t)$ is applied across the series combination of a $0.2$ henry inductance and a resistance of $100 o h m$ . Calculate:
(i) phase difference between the current and the applied voltage
(ii) average power consumed in the circuit.