Question

In a series resonant $LCR$ circuit, the voltage across $R$ is $100V$ and $R=1k\Omega$ with $C=2\mu F$. The resonant frequency $\omega$ is $200rad{s}^{-1}$. At resonance the voltage across $L$ is:

• A. $40V$
• B. $250V$
• C. $4\times {10}^{-3}V$
• D. $2.5\times {10}^{-2}V$

Answer 1

The correct option is B

$250V$

Step 1: Given data

Voltage across the resistor,$V=100V$

Resistance,$R=1k\Omega$

Capacitance, $C=2\mu F$

Resonant frequency, $\omega =200rad{s}^{-1}$

Step 2: Formula used

We know the current is given as,

$I=\frac{E}{Z}$

where $E$ is the resultant emf and $Z$ is the impedance given by the equations,

$$\begin{gathered} E=\sqrt{V_R^2+{\left(V_L-V_C\right)}^2} \\ Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2} \end{gathered}$$

Where $V_R,V_L,V_C$ are voltages across resistance, inductor and capacitor respectively and $R,X_L,X_C$ are resistance, inductive reactance and capacitive reactance respectively.

Step 3: Calculating the inductive reactance

We know, at resonance

$X_L=X_C$ and

$V_L=V_C$

Here,

$$\begin{gathered} X_L=X_C=\frac{1}{C\omega } \\ \Rightarrow X_L=\frac{1}{2\times 200\times {10}^{-6}} \\ \Rightarrow X_L=2500\to \left(1\right) \end{gathered}$$

Therefore,

$Z=R$ and

$E=V_R$

Step 4: Calculating the current

Current,

$$\begin{gathered} I=\frac{V_R}{R} \\ \Rightarrow I=\frac{100}{{10}^3} \\ \Rightarrow I=0.1A\to \left(2\right) \end{gathered}$$

Step 5: Calculating the voltage across the inductor

We know,

$$\begin{gathered} V_L=X_L\times I \\ \Rightarrow V_L=2500\times 0.1(from(1)and(2) \\ \Rightarrow V_L=250V \end{gathered}$$

The voltage across $L$ at resonance is equal to $250V$

Hence, option (B) is the correct option.