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Question

When 100V D.C. is applied across a coil, a current of 1A flows through it. When 100V A.C. of frequency 50Hz is applied to the same coil only 0.5A current flows through it. Calculate the resistance, impedance, and self-inductance of the coil?


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Solution

Step 1: Given data

Current in D.C, I=1A

Current in A.C, I=0.5A

Frequency = 50Hz

100V D.C. is applied across a coil.

100V A.C. is applied to the same coil.

Step 2: Formula used

R=VI

Here, R is the resistance, I is current, and V is the voltage or potential difference.

Z=R2+(ωL)2

Here, R is the resistance, Z is the impedance, and L is the inductance.

Step 3: Determine the resistance

In the case of a coil L-R circuit

I=VZ, Z=R2+(ωL)2

When DC is applied,

R=VIR=1001R=100Ω

Here,

R is the resistance.

I is the current.

V is the voltage or potential difference.

Step 3: Calculate the Impedance:

When AC of 50Hz is applied

I=VZ,

Z=VI =1000.5=200ohm

Step 4: Find the self-inductance of the coil

By using the impedance formula, we get

Z=R2+(ωL)2

Now,

ωL2=Z2-R2

Here,

R is the resistance.

Z is the impedance.

L is the inductance.

We know that, ω=2πf then,

(2πfL)2=2002-1002 =3×10-4

Then,

Self-inductance will be:

L=3×1022π×50L=3πL=0.55H

Therefore, the resistance, impedance, and self-inductance of the coil are R=100Ω,Z=200Ω,L=0.55H respectively..


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