Question

A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit
in which R=30, L = 25.48 mH and C = 796 uf. Find a) The impedance of the circuit and
b) The phase difference between the voltage across source and the currents.​

Answer 1

Answer:

Impedance  = 30.27 Ω

Phase difference  = tan⁻¹ (2/15)

Explanation:

Given:

• Peak value = 283 V
• Frequency = 50 Hz
• R = 30 Ω
• L = 25.58 mH
• C = 796 μF

To Find:

• Impedance of the circuit
• Phase difference between the voltage across source and the current

Solution:

First finding the inductive resistance and capacitive resistance.

We know that,

$\tt X_L=\omega\: L$

where L is the inductance and ω = 2 π ν

Substitute the data,

$\tt X_L=2\: \pi \times 50\times25.48\times 10^{-3}$

$\implies \tt 8\: \Omega$

Now finding the capacitive resistance given by,

$\tt X_C=\dfrac{1}{\omega\:C}=\dfrac{1}{2\times \pi\times \nu\times C}$

Substitute the data,

$\tt X_C=\dfrac{1}{2\times \pi \times 50\times 796\times 10^{-6}}$

$\implies \tt 4\: \Omega$

Now the impedance of the circuit is given by,

$\tt Z=\sqrt{R^2+(X_L-X_C)^2}$

Substitute the data,

$\tt Z=\sqrt{30^2+(8-4)^2}$

$\tt Z=\sqrt{900+16}$

$\implies \tt \sqrt{916} =30.27\: \Omega$

Now finding the phase difference,

$\tt tan\: \phi=\dfrac{X_L-X_C}{R}$

Substitute the data,

$\tt tan\: \phi=\dfrac{8-4}{30}$

$\tt tan\: \phi=\dfrac{2}{15}$

$\tt \phi=tan^{-1}\bigg(\dfrac{2}{15}\bigg)$