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Question
A circuit is set up by connecting inductance L = 100 mH, resistor R = 100 Ω and a capacitor of reactance 200 Ω in series. An alternating emf of \[150\sqrt{2}\] V, 500/π Hz is applies across this series combination. Calculate the power dissipated in the resistor.
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Solution
Given the r.m.s value of emf in the circuit Ev = 150√2 V
The impedance of the LCR- circuit is given by
\[Z = \sqrt{R^2 + \left( X_L - X_C \right)^2}\]
\[ = \sqrt{R^2 + \left( 2\pi fL - 200 \right)^2}\]
\[ = \sqrt{{100}^2 + \left( 2\pi \times \frac{500}{\pi} \times 0 . 1 - 200 \right)^2}\]
\[ = 100\sqrt{2} \Omega\]
The r.m.s value of current Iv in the circuit is given by
\[I_v = \frac{E_v}{Z} = \frac{150\sqrt{2}}{100\sqrt{2}} = 1 . 5 A\]
Power dissipated in the resistor = EvIv = 318.2 W
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