GATE EE 2007 Electrical Circuits Q 2 - Grad Plus
GATE EE 2007 Electrical Circuits Q 2

GATE EE 2007 Electrical Circuits Q 2

In the figure given below all phasors are with reference to the potential at point ”O”. The locus of voltage phasor Vyx as R is varied from zero to infinity is shown by

(a)
(b)
(c)
(d)

Ans. (a)

Explanation:

Assuming capacitive reactance as, Xc

I=\frac{V\angle0^\circ\;+\;V\angle0^\circ}{R\;-jX_C} I=\;\frac{2V}{R-jX_c}

Using KVL in upper loop, we have,

Vyx + IR – V =0

V_{yx\;}=V-IR V_{yx}\;=\;V-\left(\frac{2V}{R-jX_c}\right)R V_{yx}=\frac{V(R\;-\;jX_c)\;-2VR}{R\;-\;jX_c} V_{yx}=\frac{V(R\;+jX_c)}{(R\;-jX_c)} V_{yx}\;=\;-V\left[\frac{R\;+\;jX_c}{R\;-\;jX_c}\right]

Now, let us vary R from 0 to infinity.

For, R = 0

V_{yx}\;=\;-V\left[\frac{0\;+\;jX_c}{\;0-\;jX_c}\right]\;=V

For, R approaching to infinity, let us re-write the equation VYX

V_{yx}\;=\;-V\left[\frac{R\;+\;jX_c}{R\;-\;jX_c}\right] V_{yx}\;=\;-V\times\;\left[\frac{1\;+\;j{\displaystyle\frac{X_c}R}}{\;1-\;j\;{\displaystyle\frac{X_C}R}}\right]\;

For, Vyx approaching ∞, we have Vyx = -V

Hence, magnitude of Vyx is V. And, hence options (c) and (d) are not possible.

And with the given options (a) is correct one.

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