Question

A supply of 250v, 50hz is applied to a series rc circuit.If power absorbed by resistor be 400w at 160v. The value of capacitor c will be

Answer 1

Answer:

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Answer 2

Answer:

Required capacitor is 41.5 μF.

Explanation:

Given, 250 V, 50 hz is applied to a series rc circuit and power absorbed by resistor be 400 W at 160 V.

Now we want to find capacitor C.

We know ,

$V_C = \sqrt{ {V_s}^{2} - V}$

$V_C = \sqrt{ {250}^{2} - 160}$

$V_C = \sqrt{6900} = 192.09≈192 \: V$

We know that,

$P_R = \frac{V_R}{R}$

$P_R = 400 W$(given)

$R = \frac{ {V_R}^{2} }{P_R} = 64$

Current through circuit will be

$I = \frac{V_R }{R } = \frac{160}{64} = 2.54$

Now Z$= \frac{ |V_s | }{ I} = \frac{250}{2.5} = 10$

Total impedance of the circuit is

$Z = \sqrt{ { R }^{2} + | X_C| }$

So,

$|X_C | = \sqrt{ {Z }^{2} - { R}^{2} } = 76.8$

Now,

$\frac{1}{wC} = 76.8$

$C = \frac{1}{w \times 76.8} = \frac{1}{2\pi \times 50 \times 76.8} = 41.5$