Question

In series LCR circuits the voltages across inductor, capacitor and resistance are $30\,V$, $30\,V$ and $60\,V$ respectively. Find the net emf of the circuit.

Answer 1

The net emf of the LCR series circuit is determined by using the emf formula, the emf of the LCR series circuit depends on the voltage across the inductor, voltage across the capacitor and the voltage across the resistance. By using this, the net emf can be determined.

Formula Used:
The emf of the LCR series circuit is given by,
$e = \sqrt {{{\left( {{V_L} - {V_C}} \right)}^2} + V_R^2}$
Where, ${V_L}$ is the voltage across the inductor, ${V_C}$ is the voltage across the capacitor, ${V_R}$ is the voltage across the resistance and $e$ is the net emf of the circuit.

Complete step by step answer:
Given that,
The voltage across the inductor is, ${V_L} = 30\,V$
The voltage across the capacitor is, ${V_C} = 30\,V$
The voltage across the resistance is, ${V_R} = 60\,V$
Now,
The emf of the LCR series circuit is given by,
$e = \sqrt {{{\left( {{V_L} - {V_C}} \right)}^2} + V_R^2} \,...................\left( 1 \right)$
By substituting the voltage across the inductor, voltage across the capacitor and the voltage across the resistance in the above equation (1), then the above equation (1) is written as,
$e = \sqrt {{{\left( {30 - 30} \right)}^2} + {{60}^2}}$
By subtracting the terms in the above equation, then the above equation is written as,
$e = \sqrt {{0^2} + {{60}^2}}$
By adding the terms in the above equation, then the above equation is written as,
$e = \sqrt {{{60}^2}}$
By taking the square in the above equation, then the above equation is written as,
$e = \sqrt {3600}$
By taking the square root on the above equation, then the above equation is written as,
$e = 60\,V$
Thus, the above equation shows the net emf of the LCR series circuit.

Note: The emf of the LCR series circuit is equal to the square root of the sum of the square of the voltage across resistance and the square of the difference of the voltage across the inductor and the voltage across the capacitor.