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In a series RL circuit with inductive reactance X_L = 5 Ohm, resistance R = 5 Ohm, and supply voltage V = 100*sqrt(2) * sin(omega*t) volts, find the potential difference across the inductor at the instant when the potential difference across the source becomes half its maximum value.

1. 96.5 V
2. 100 V
3. 50*sqrt(2) V
4. 25*sqrt(2) V

Correct answer: 50*sqrt(2) V

Solution

Z = sqrt(5² + 5²) = 5*sqrt(2) Ohm. Peak current I0 = 100*sqrt(2)/(5*sqrt(2)) = 20 A. The voltage across the source at instant t: Vₛ(t) = 100*sqrt(2)*sin(omega*t). Vₛ = V_max/2 => sin(omega*t) = 1/2. The instantaneous current i(t) = I0*sin(omega*t - phi) where phi = arctan(X_L/R) = arctan(1) = 45 deg. At sin(omega*t)=1/2, omega*t = pi/6, so i = 20*sin(pi/6 - pi/4) = 20*sin(-pi/12). V_L = i * X_L = 20*5*sin(-pi/12)... this gives a negative value. Re-examining: V_L(t) = I0*X_L*sin(omega*t - phi + pi/2) = 20*5*cos(omega*t - pi/4). At omega*t = pi/6: V_L = 100*cos(pi/6 - pi/4) = 100*cos(-pi/12) = 100*cos(15 deg) ≈ 96.6 V ≈ 96.5 V.

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