Question

In the series LCR circuit, the voltmeter and ammeter readings are:

A) $V = 100\,V$, $I = 2\,A$
B) $V = 100\,V$, $I = 5\,A$
C) $V = 1000\,V$, $I = 2\,A$
D) $V = 300\,V$, $I = 1\,A$

Answer 1

Hint: In LCR circuit, if the resistor, inductor and capacitor is connected in series, then the voltmeter and ammeter readings can be determined by using the current and voltage formula. By using the voltage formula, the voltmeter reading can be determined, and by using the current formula the ammeter reading can be determined.

Formula used:
The expression for finding the reading of voltmeter is
$V = \sqrt {{V_R}^2 + {{\left( {{V_L} - {V_C}} \right)}^2}} $
Where, $V$ is the voltmeter reading, ${V_R}$ is the voltage across resistor, ${V_L}$ is the voltage across the inductor, ${V_C}$ is the voltage across the capacitor.
By ohm’s law,
$V = IR$
Where, $V$ is the voltage, $I$ is the current and $R$ is the resistance.

Complete step by step solution:
Given that,
Resistance, $R = 50\,\Omega $,
Voltage across inductor, ${V_L} = 400\,V$
Voltage across capacitor, ${V_C} = 400\,V$
Voltage across resistor, ${V_R} = 100\,V$
The expression for finding the reading of voltmeter is
$V = \sqrt {{V_R}^2 + {{\left( {{V_L} - {V_C}} \right)}^2}} \,.................\left( 1 \right)$
On substituting the voltage across inductor, capacitor and resistor in the above equation (1), then
$V = \sqrt {{{\left( {100} \right)}^2} + {{\left( {400 - 400} \right)}^2}} $
By simplifying the terms, then the above equation is written as,
$V = \sqrt {{{\left( {100} \right)}^2} + 0} $
The above equation is written as,
$V = \sqrt {{{\left( {100} \right)}^2}} $
In the above equation the square and the square root get cancel each other, then the above equation is written as,
$V = 100\,V$
Thus, the above equation shows the voltage reading shown by the voltmeter.
Now,
By using the ohm’s law,
$V = IR\,............\left( 2 \right)$
We have to find the current, so keep the current in one side and the other terms in other side, then the above equation is written as,
$I = \dfrac{V}{R}$
Now, substituting the voltage value and the resistance value in the above equation, then the above equation is written as,
$I = \dfrac{{100}}{{50}}$
On dividing, then
$I = 2\,A$
Thus, the above equation shows the current reading shown by the ammeter.

Hence, the option (A) is the correct answer.

Note: The voltage across the inductor and capacitor is same, by subtracting these terms it will become zero, in equation (2), the voltage value substituted is the applied potential difference to the LCR circuit and it is mentioned in the circuit diagram. Then by substituting, the current reading shown by ammeter is determined.