Question

In the given circuit, the readings of voltmeters $V_{1}$ and $V_{2}$ are 300 V each. The reading of voltmeter $V_{3}$ and ammeter A are respectively

• A. 100 V, 2.0 A
• B. 150 V, 2.2 A
• C. 220 V, 2.0 A
• D. 220 V, 2.2 A

Hint:

When $V_{L} = V_{C}$, the circuit is in resonance! The entire source voltage drops purely across the resistor ($V_{R} = V_{\text{source}}$), and the impedance becomes minimum ($Z = R$).

Answer 1

The Correct Option is D

Step 1: Series LCR voltage rule.
In a series LCR circuit the supply voltage is $V = \sqrt{V_{R}^{2} + (V_{L} - V_{C})^{2}}$.

Step 2: Spot resonance.
$V_{1}$ reads $V_{L}=300$ V and $V_{2}$ reads $V_{C}=300$ V. Since they are equal, $V_{L}=V_{C}$, so the circuit is at resonance.

Step 3: Find $V_{3}=V_{R}$.
At resonance $V_{L}-V_{C}=0$, so $V = V_{R}$. With supply $220$ V, the resistor voltage $V_{3} = 220$ V.

Step 4: Impedance at resonance.
At resonance the impedance equals the resistance, $Z = R = 100\,\Omega$.

Step 5: Find the current.
\[ I = \frac{V}{Z} = \frac{220}{100} = 2.2 \text{ A} \]
Step 6: State both readings.
So $V_{3} = 220$ V and the ammeter reads $2.2$ A, which is option 4.
\[ \boxed{V_{3} = 220 \text{ V},\ I = 2.2 \text{ A}} \]