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)

Short answer

(a) Peak current through inductor is $0·79A$

(b) Peak current through inductor is$0·00079A$

Step-by-step solution

Given information

We have given that,

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

Peak voltage is,$V_0=10V$

Part (a) Step 1: Given information

We have given that,

Frequency is,$f_1=100Hz$

Part (a) Step 2: Simplify

Let resistance of inductor is, $X_L$

We know that in inductive circuit resistance is given as,

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

By ohm's law, we know that,

role="math" localid="1649883173982" $$\begin{gathered} Current=\frac{Voltage\left(V_0\right)}{Resis\tan ce\left(X_L\right)} \\ Current=\frac{10}{12·57} \\ Current=0·79A \end{gathered}$$

Hence, Current through inductor is$0·79A$

Part (b) Step 1: Given information 

We have given that,

Frequency is,$f_2=100kHz=100\times {10}^{3}Hz$

Part (b) Step 2: Simplify

Let resistance of inductor is,$X_L$

We know that in inductive circuit resistance is given as,

$$\begin{gathered} X_L=2\times \mathrm{\pi }\times {\mathrm{f}}_2\times \mathrm{L} \\ {\mathrm{X}}_{\mathrm{L}}=2\times 3·14\times 100\times {10}^3\times 20\times {10}^{-3} \\ {\mathrm{X}}_{\mathrm{L}}=12566\mathrm{\Omega } \end{gathered}$$

By ohm's law, we know that,

Hence, Current through inductor is$0.00079$