A coil when connected across a steady DC voltage V draws a current I₀ . When it is connected across an AC source of the same rms voltage V and frequency f , it draws an rms current of I₀ 2 . If the frequency of the AC source is increased to 2f while keeping the rms voltage constant, what will be the new rms current drawn by the coil?

Step-by-step solution

When connected to a steady DC voltage V , the inductive reactance is zero. The resistance of the coil is R = V I₀ . When connected to an AC source of frequency f , the impedance is Z = V I_ rms = V I₀ 2 = 2 R . Since Z = R^2 + X_L^2 , we have 2 R = R^2 + X_L^2 , which gives X_L = R . When the frequency is doubled to 2f , the new inductive reactance becomes X_L' = 2X_L = 2R . The new impedance is Z' = R^2 + (2R)^2 = 5 R . The new rms current drawn is I' = V Z' = V 5 R = I₀ 5 . Answer: I₀ 5