Question

A resistor of 100 ohm,a pure inductance coil of L=0.5H and capacitor are in series in a circuit containing an account source of 200V,50Hz.In the circuit current is ahead of voltage by 30 degree.Find the value of the capacitance

Answer 1

Answer:

From the given data:

L=0.1H

R=10

Power factor i.e cos∅=1, cos∅=R/Z (Z=impedance)

therefore, 1=10/z

z=10ΩΩ.

XL(inductive reactance= ωL

=2πfL= 10π≈ 31.4Ω

Z=√R²+(XL-XC)²

sub. all the values

You will get Capacitance ∵ XC as 98.5mC

I= V/R

=220/10

=22A

Hope it was helpful

Regards

adarsh

Answer 2

The value of the capacitance is 1.46 F.

Explanation:

The current leads voltage by angle is given as

$tan\theta=\frac{X_C-X_L}{R}$

Here, $X_C$ is the capacitor reactance and $X_L$ is the Inductive reactance.

In the circuit the capacitor reactance,

$X_C=\frac{1}{2\pi fC}$

Here, f is the frequency and C is the capacitance.

In the circuit the inductive reactance,

$X_L=2\pi f L$

So, the above equation becomes,

$tan\theta=\frac{\frac{1}{2\pi fC}-2\pi f L}{R}$

Given f = 50 Hz, R = 100 ohm, L = 0.5 H and  $\theta=30^0$.

Substitute the all the given values in above equation, we get

$tan30=\frac{\frac{1}{2\pi \times50Hz\times C}-2\pi\times50Hz\times0.5H}{100\Omega}$

After solving we get

C =1.46 F.

Thus, the value of the capacitance is 1.46 F.

#Learn More: inductive reactance, capacitor reactance.

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