A 20mHinductor 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

Expert verified

(a) Peak current is0.79A

(b) Peak current is0.00079A

Step by step solution

01

Part(a)Step 1: Given information

We have been given that,

Inductance of inductor is, L=20mH=20×10-3H

Peak Voltage is, V0=10V and frequency f1=100Hz

02

Part (a) Step 2: Simplify

Let inductive reactance is,XL

We know that in inductive circuit reactance is given as,

localid="1649932325839" XL=2π×f1×LXL=2×3·14×100×20×10-3XL=12·57Ω

By ohm's law, we know that,

Current,I=VoXLI=1012·57I=0·79A

Hence, Current is0.79A

03

Part (b) Step 1: Given information

We have been given that,

Frequency is,f1=100kHz=100×103Hz

04

Part (b) Step 2: Simplify

Let inductive reactance is,XL

We know that in inductive circuit reactance is given as,

localid="1649932419844" XL=2π×f1×LXL=2×3·14×100×103×20×10-3XL=12566Ω

By ohm's law, we know that,

Current,I=VoXLI=1012566I=0·00079A

Hence, Current is0.00079A0·00079A

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