Question

A circuit operating at 360/(2pi) Hz contains a 1muF capacitor and a 20 Omega. resistor. How large an inductor must be added in series to make the phase angle for the circuit zero? Calculate the current in the circuit if the applied voltage is 120 V.

Answer 1

7.72 H inductor must be added in series to make the phase angle for the circuit 0. The current in the circuit is 6 A.

• In LCR circuit the current and voltage are in phase at the condition of resonance only.
• Given: Frequency = 360/2pi; Capacitance (C) = 1 muF; Resistance (R) = 20 omega; Voltage (V) = 120 V.
• Angular frequency (w) = 360.
• Using Resonance Condition:
• Inductance (L) = 1/(w^2)C = 1/360*360*10^-6 = 7.72 H
• At Resonance Reactance of capacitance and inductor cancel out.
• Current (I) = V/R = 120/20 = 6 A

Answer 2

Given :

A circuit operating at  $\dfrac{360}{2\pi }$ Hz

The value of capacitor = C = 1 $\mu$F

The value of resistor = R = 20  ohm

Applies voltage = 120 volt

To Find :

The value of inductor added in series

Solution :

In LRC circuit  phase angle is zero only when the circuit is at resonance condition .

And  At resonance condition

Inductive Reactant = capacitance Reactant

So,  $X_L$  = $X_C$

i.e  L $\omega$ = $\dfrac{1}{C\omega }$                ...........1

∵     $\dfrac{1}{f}$  = $\dfrac{2\pi }{\omega }$

Or,  $\dfrac{2\pi }{360}$  = $\dfrac{2\pi }{\omega }$            ( given )

∴     ω = 360   Hz

So, From eq 1

   L × 360 = $\dfrac{1}{1 \times 10^{-6}\times 360 }$

Or,     L × 360 = $\dfrac{10^{6} }{360}$

Or,      L = $\dfrac{10^{6} }{360^{2} }$

∴   Inductance = L = 7.71  Henry

Again

Since, At resonance condition, circle is purely Resistive

So, Current flowing through the circuit = I

i.e    I = $\dfrac{voltage}{resistance}$

Or,   I = $\dfrac{120}{20}$

∴   Current = I = 6 Ampere

Hence,

The inductor of value must be added in series to make the phase angle for the circuit zero is  1.71 Henry

And, The current flowing in the circuit is 6 Ampere   Answer