Question

In the circuit shown in the figure, the current measured by the ammeter is

• A. $1A$
• B. $2A$
• C. $3A$
• D. $0$

Answer 1

The correct option is C

$3A$

Step 1: Series $LCR$ Circuit

We know that the current in a series $LCR$ circuit impressed by an ac emf $v=V_0\cos \omega t$ is given by

$i=I_0\cos \left(\omega t-\varphi \right)$,

Where $I_0=\frac{V_0}{\sqrt{R^2+{\left(\omega L-{\displaystyle \frac{1}{\omega C}}\right)}^2}}$ and $\varphi ={\tan }^{-1}\frac{\omega L-{\displaystyle \frac{1}{\omega C}}}{R}$

The RMS current is $I_r=\frac{V_r}{\sqrt{R^2+{\left(\omega L-{\displaystyle \frac{1}{\omega C}}\right)}^2}}$, $\left(1\right)$

Where $V_r$ is the RMS voltage. Thus, $I_r$ depends on the frequency $\omega$ of the impressed ac emf.

Step 2: Phasor Diagram

The given circuit can be represented in a phasor diagram as

Step 3: Calculate the Current

From the graph we get

$\sqrt{{\left(100-60\right)}^2+{V_R}^2}=V_{net}$

$\Rightarrow \sqrt{{\left(40\right)}^2+{V_R}^2}=50$

$\Rightarrow {\left(40\right)}^2+{V_R}^2={\left(50\right)}^2$

$\Rightarrow {V_R}^2=2500-1600$

$\Rightarrow {V_R}^2=900$

$\Rightarrow V_R=30$

Given that $R=10\Omega$

$\therefore I=\frac{V}{R}$

$=\frac{30}{10}=3A$

Hence, the correct answer is option (C).