An alternating emf given by the equation E = | vedclass

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(D) The given emf is $E = 200 \sin(50 \pi t)$.At $t = 1 \ s$,the instantaneous voltage is $E = 200 \sin(50 \pi 	imes 1) = 200 \sin(50 \pi) = 0 \ V$.H

Vedclass · Accepted Answer

$307$

Vedclass · Answer

(D) The given emf is $E = 200 \sin(50 \pi t)$.At $t = 1 \ s$,the instantaneous voltage is $E = 200 \sin(50 \pi 	imes 1) = 200 \sin(50 \pi) = 0 \ V$.However,the power dissipated by a resistor in an $LR$ circuit is given by $P = I^2 R$.The impedance of the circuit is $Z = \sqrt{R^2 + X_L^2} = \sqrt{30^2 + 40^2} = 50 \ \Omega$.The current in the circuit is $I = I_0 \sin(50 \pi t - \phi)$,where $I_0 = E_0 / Z = 200 / 50 = 4 \ A$.The phase angle $\phi$ is given by $	an \phi = X_L / R = 40 / 30 = 4/3$,so $\phi = 53^{\circ}$.At $t = 1 \ s$,the instantaneous current is $I = 4 \sin(50 \pi - 53^{\circ}) = 4 \sin(-53^{\circ}) = -4 \sin(53^{\circ}) = -4 	imes 0.8 = -3.2 \ A$.The instantaneous power dissipated by the resistor is $P = I^2 R = (-3.2)^2 	imes 30 = 10.24 	imes 30 = 307.2 \ W$.Thus,the power is close to $307 \ W$.

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