Answer to Question #126680 in Physics for Susan Williams

Question

An inductor, L, of 2 H and a resistor, R, of 1000Ω are in series across an a.c. voltage of 10 V, frequency 60 Hz. Calculate
(i) the impedance in the circuit

(ii) the current in r.m.s. flowing,

(iii) the r.m.s voltage across the inductor.

Answer 1

(i) The impedance of a series circuit is:

$$Z = \sqrt{R^2 + (2\pi fL)^2} = \sqrt{1000^2 + (2\pi \cdot 60\cdot 2)^2} \approx1252.4\Omega$$

(ii) the current in r.m.s. flowing, according to the Ohm's law:

$$I = \dfrac{V}{Z} = \dfrac{10V}{1252.4\Omega} \approx 0.00798A =7.98mA$$

(iii) the r.m.s voltage across the inductor according to the Ohm's law:

$$V_L = IZ_L = I\cdot 2\pi fL = 0.00798\cdot 2\pi\cdot 60\cdot 2 \approx 6.02V$$

Answer. (i)1252.4 Ohm, (ii)7.98 mA, (iii)6.02 V.