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  • A resistor R and an inductor L are connected in series to a source . Find the (a) peak value of the voltage drops

A resistor R and an inductor L are connected in series to a source V = V_{0} \sin \omega t.

Find the (a) peak value of the voltage drops across R and across L, (b) phase difference between the applied voltage and current. Which of them is ahead?

 

 

 
 
 
 
 

Answers (1)

The given circuit is RL series circuit.

The peak value of current through the circuit is 

I_0=\frac{V_0}{Z}=\frac{V_0}{\sqrt{R^2+X_L^2}}

a)

The peak value of voltage across R

=I_0R=\frac{V_0R}{\sqrt{R^2+X_L^2}}

The peak value of voltage across L

=I_0X_L=\frac{V_0X_L}{\sqrt{R^2+X_L^2}}

b) The phase difference

\phi=tan^{-1}\frac{X_L}{R}

Here the voltage leads the current by \phi

Posted by

Safeer PP

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