Question

A $20mH$inductor is connected across an AC generator that produces a peak voltage of $10V$. What is the peak current through the inductor if the emf frequency is (a) $100Hz$? (b) $100kHz$?

Short answer

(a) Peak current is$0.79A$

(b) Peak current is$0.00079A$

Step-by-step solution

Part(a)Step 1: Given information

We have been given that,

Inductance of inductor is, $L=20mH=20\times {10}^{-3}H$

Peak Voltage is, $V_0=10V$ and frequency $f_1=100Hz$

Part (a) Step 2: Simplify

Let inductive reactance is,$X_L$

We know that in inductive circuit reactance is given as,

localid="1649932325839" $$\begin{gathered} X_L=2\mathrm{\pi }\times {\mathrm{f}}_1\times \mathrm{L} \\ {\mathrm{X}}_{\mathrm{L}}=2\times 3·14\times 100\times 20\times {10}^{-3} \\ {\mathrm{X}}_{\mathrm{L}}=12·57\mathrm{\Omega } \end{gathered}$$

By ohm's law, we know that,

$$\begin{gathered} Current,I=\frac{V_o}{X_L} \\ I=\frac{10}{12·57} \\ I=0·79A \end{gathered}$$

Hence, Current is$0.79A$

Part (b) Step 1: Given information

We have been given that,

Frequency is,$f_1=100kHz=100\times {10}^{3}Hz$

Part (b) Step 2: Simplify

Let inductive reactance is,$X_L$

We know that in inductive circuit reactance is given as,

localid="1649932419844" $$\begin{gathered} X_L=2\mathrm{\pi }\times {\mathrm{f}}_1\times \mathrm{L} \\ {\mathrm{X}}_{\mathrm{L}}=2\times 3·14\times 100\times {10}^3\times 20\times {10}^{-3} \\ {\mathrm{X}}_{\mathrm{L}}=12566\mathrm{\Omega } \end{gathered}$$

By ohm's law, we know that,

$$\begin{gathered} Current,I=\frac{V_o}{X_L} \\ I=\frac{10}{12566} \\ I=0·00079A \end{gathered}$$

Hence, Current is$0.00079A$$0·00079A$