Question

In a series LCR circuit at resonance, the applied ac voltage is $220\,V$.The potential drop across the inductance is $110V$. What is the potential drop across the resistance at resonance?

Answer 1

Hint: The reduction in electrical potential along the path of a current flowing in an electrical circuit is known as voltage drop. Since any of the energy supplied is dissipated, voltage decreases in the source's internal resistance, and across conductors are undesirable.

Complete step by step answer:
A resistor of $R$ ohms, a capacitor of $C$ farad, and an inductor of $L$ Henry are all connected in a series combination in an electronic LCR circuit.
Resistance: Resistor is an electrical device that decreases the amount of current flowing through it.
Capacitor: Capacitor is a device for storing electrical energy that consists of two shielded conductors in close proximity. The parallel-plate capacitor is a simple example of such a storage device.
Inductor: An inductor is a passive electronic component that can store electrical energy as magnetic energy.

From the question, we know that the applied ac voltage, $V = 220V$ , and the voltage across the inductor, ${V_L} = 110V$
At resonance, the voltage across an inductor is equivalent to the voltage across the capacitor
$ \Rightarrow {V_L} = {V_C}$
Therefore, ${V_C} = 110V$
To find the voltage drop across the resistance, we use this formula:
$V = \sqrt {V_R^2 + {{\left( {{V_L} - {V_C}} \right)}^2}} $
Where ${V_R}$ represents the voltage across the resistor, ${V_L}$​ represents the voltage across the inductor and ${V_C}$​ represents the voltage across the capacitor.

 Now we substitute the values in the above formula,
$V = \sqrt {V_R^2 + {{\left( {{V_L} - {V_C}} \right)}^2}} $
On squaring both the sides, we get,
${V^2} = V_R^2 + {\left( {{V_L} - {V_C}} \right)^2}$
On rearranging the terms, we get,
$V_R^2 = {V^2} - {\left( {{V_L} - {V_C}} \right)^2} \\
\Rightarrow V_R^2 = {220^2} - {\left( {110 - 110} \right)^2} \\
\Rightarrow V_R^2 = {220^2}$
Now, we shall take square root on both sides,
$\therefore {V_R} = 220\,V$

Hence, the potential drop across the resistance is equal to $220\,V$.

Note: The LCR circuits are used to track narrow-band frequencies in a wide spectrum of radio waves. The LCR circuit is used to tune AM/FM radio frequencies. Depending on the frequency, it may be used as a low pass, band-pass, high pass, or band-stop filter.