GATE EE 2005 Electrical Circuits Q 3 - Grad Plus
GATE EE 2005 Electrical Circuits Q 3

GATE EE 2005 Electrical Circuits Q 3

The RL circuit of the figure is fed from a constant magnitude, variable frequency sinusoidal voltage source Vin . At 100Hz, the R and L elements each have a voltage drop URMS . If the frequency of the source is changed to 50HZ, then new voltage drop across R is

a) \sqrt{\frac58}U_{rms}

b)\sqrt{\frac23}U_{rms}

c)\sqrt{\frac85}U_{rms}

d)\sqrt{\frac32}U_{rms}

Ans. (c)

Explanation:

At, f = 100Hz, \left|V_R\right|\;=\left|V_L\right|

Let the current through the circuit is I. And as series conncetion current through both R and L will be same.

As, \left|V_R\right|\;=\left|V_L\right|

We have, IR\;=\;IX_L=\;I\omega L

Hence, R=X_L=\omega L …(1)

Now, current is given as, I = Vin/Z

I=\;\frac{V_{in}}{\sqrt{R^2\;+X_L^2}}\;=\frac{V_{in}}{\sqrt{R^2\;+R^2}}\;=\frac{V_{in}}{\sqrt2R}

Voltage drop across resistance is given as,

V_R\;=\left(\frac{V_{in}}{\sqrt2R}\right)\;\times R\;=\frac{V_{in}}{\sqrt2}

But given that, V_R=U_rms

Hence, V_{in}\;=\sqrt2\;U_{rms}

Now, resistance (R) is independant of frequency. But, inductive reactance (XL) depends on the frequency.

X_L=\omega L=2\pi fL X_L\propto f

Now frequency of source is changed from 100Hz to 50 Hz, hence frequency is halved. Hence, new reactance would also be halved.

X_{L}^{'}=\frac{X_L}2

Hence from equation(1), X_{L}^{'}=\frac R2

Hence, new current I’ is given as

I'\;\;=\frac{V_{in}}{\sqrt{R^2\;+{(X_L')}^2}} \;\;=\frac{V_{in}}{\sqrt{R^2\;+\left({\displaystyle\frac R2}\right)^2}} \;\;=\frac{2V_{in}}{\sqrt5R} \;V_R'\;\;=I'R\;=\left(\frac{2V_{in}}{\sqrt5R}\right)R\;=\frac2{\sqrt5}V_{in}

But, V_{in}\;=\sqrt2\;U_{rms}

\;V_R'\;\;=\;\frac2{\sqrt5}\times\left(\sqrt2U_{rms}\right) =\;\frac{2\sqrt2}{\sqrt5}U_{rms=\;}\sqrt{\frac85}U_{rms}
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