Question

An alternating power supply of 220V is applied across a series circuit of resistance$10\sqrt 3 \Omega$, capacitive reactance $40\Omega$ and inductive reactance$30\Omega$. The respective current in the circuit for zero and infinite frequencies are
A. $2A,\dfrac{1}{2}A$
B. $0A,10A$
C. $10A,0A$
D. $0A,0A$

Answer 1

In this question, we need to determine the current in the circuit for zero and infinite frequencies. Since the value of alternating voltage and the resistance, capacitive reactance and inductive reactance are given, first we will find the impedance of the circuit and then we will find the current in the circuit when the frequency is zero and infinite.

Formula used:
Impedance of a series resonating circuit is given as $Z = \sqrt {{R^2} + {{\left( {{X_C} - X{}_L} \right)}^2}}$ where,
R is the resistance in the circuit, ${X_c}$ is the capacitive reactance in the circuit and ${X_L}$ is the inductive reactance in the circuit.

Complete step by step answer:
Let the frequency of the circuit be $\omega$
The alternating power supply ${v_0} = 220V$
Resistance $R = 10\sqrt 3 \Omega$
Capacitive reactance $= \dfrac{1}{{\omega C}} = 40\Omega$
Inductive reactance $= \omega L = 30\Omega$
Now since the resistance, inductive reactance and the capacitive reactance are in the series so the total impedance of the circuit will be
$Z = \sqrt {{R^2} + {{\left( {{X_C} - X{}_L} \right)}^2}}$
This can be also written as
$Z = \sqrt {{R^2} + {{\left( {\dfrac{1}{{\omega C}} - \omega L} \right)}^2}}$
So the current flowing in the circuit containing resistor, capacitor and inductor will be equal to
${i_0} = \dfrac{{{v_0}}}{Z}$
Hence by substituting the value of the voltage and the impedance we can write current

$${i_0} = \dfrac{{220}}{{\sqrt {{R^2} + {{\left( {{X_C} - X{}_L} \right)}^2}} }} \\\ = \dfrac{{220}}{{\sqrt {{R^2} + {{\left( {\dfrac{1}{{\omega C}} - \omega L} \right)}^2}} }} \\\$$

Case i: When frequency of the circuit is zero,$\omega = 0$
So the current in the circuit will be

$${i_0} = \dfrac{{220}}{{\sqrt {{R^2} + {{\left( {\dfrac{1}{{\omega C}} - \omega L} \right)}^2}} }} \\\ = \dfrac{{220}}{{\sqrt {{R^2} + {{\left( {\dfrac{1}{{0 \times C}} - 0 \times L} \right)}^2}} }} \\\ = \dfrac{{220}}{{\sqrt {{R^2} + {{\left( {\infty - 0} \right)}^2}} }} \\\ = \dfrac{{220}}{\infty } \\\ = 0A \\\$$

Hence the current flowing in the series circuit when frequency of the circuit is zero is $= 0A$
Case ii: When frequency of the circuit is infinite, $\omega = \infty$
So the current in the circuit will be

$${i_0} = \dfrac{{220}}{{\sqrt {{R^2} + {{\left( {\dfrac{1}{{\omega C}} - \omega L} \right)}^2}} }} \\\ = \dfrac{{220}}{{\sqrt {{R^2} + {{\left( {\dfrac{1}{{\infty \times C}} - \infty \times L} \right)}^2}} }} \\\ = \dfrac{{220}}{{\sqrt {{R^2} + {{\left( {0 - \infty } \right)}^2}} }} \\\ = \dfrac{{220}}{\infty } \\\ = 0A \\\$$

Hence, the current flowing in the series circuit when frequency of the circuit is infinite is $= 0A$
Therefore, current in the circuit for zero and infinite frequencies are $0A,0A$

So, the correct answer is “Option D”.

Note:
Students must note that when an AC supply has zero frequency or the infinite frequency then no current will flow through the circuit. Moreover, it is interesting to note that when the frequency is infinite then, inductive reactance acts like an open circuit and the capacitive reactance acts like a short circuit and vice versa.