Question

In 220 kV, 50 Hz system, the value of reactance and capacitance upto the location of circuit breaker is 8 $\Omega$ and 0.025 $\mu$F respectively. The value of resistance which will give damped frequency of oscillation which is one fourth of natural frequency of oscillation will be _____ $\Omega$.

• A. 521.12

Answer 1

The correct option is A 521.12
We know, natural frequency of oscillation,

$f_n=\frac{1}{2\pi }\sqrt{\frac{1}{LC}}$

$X_L=2\pi fL$

$8=2\pi \times 50\times L$

$\Rightarrow L=\frac{8}{2\pi \times 50}=0.02546H$

$\therefore f_n=\frac{1}{2\pi }\sqrt{\frac{1}{0.02546\times 0.025\times {10}^{-6}}}$

$f_n=6.31kHz$

Required damped frequency of oscillation

$=\frac{f_n}{4}=1.577kHz$
Also frequency of damped oscillation,

$f=\frac{1}{2\pi }\sqrt{\frac{1}{LC}-\frac{1}{4C^2{R}^2}}$

$1577=\frac{1}{2\pi }\sqrt{\frac{1}{0.02546\times 0.025\times {10}^{-6}}-\frac{1}{4\times {(0.025\times {10}^{-6})}^2\cdot R^2}}$

$98180021.61=1571091909-\frac{4\times {10}^{14}}{R^2}$

$\frac{4\times {10}^{14}}{R^2}=1472911887$

$R^2=2.7157\times {10}^5$
$R=521.124\Omega$