Question

What will be the reading in the voltmeter and ammeter of the circuit shown ?

• A. 90V, 2A
• B. 0V, 2A
• C. 90V, 1A
• D. 0V, 1A

Answer 1

The Correct Option is B

1. Impedance of the circuit:

The impedance ($Z$) of a series RLC circuit is given by:

$Z = \sqrt{R^2 + (X_L - X_C)^2}$

In this case, $X_L = X_C = 4\,\Omega$, so the impedance is:

$Z = \sqrt{(45\,\Omega)^2 + (4\,\Omega - 4\,\Omega)^2} = \sqrt{(45\,\Omega)^2} = 45\,\Omega$

2. Current in the circuit:

The current ($I$) in the circuit is given by Ohm's law for AC circuits:

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

where $V$ is the source voltage. Substituting the given values:

$I = \frac{90\,\text{V}}{45\,\Omega} = 2\,\text{A}$

Since the ammeter is connected in series with the circuit, it will read this current.

3. Voltage across the voltmeter:

The voltmeter is connected across the inductor and capacitor. Since $X_L$ and $X_C$ are equal, the voltage across the inductor ($V_L = IX_L$) and the voltage across the capacitor ($V_C = IX_C$) are equal in magnitude but opposite in phase. Therefore, the net voltage across the inductor and capacitor combination is:

$V_{LC} = V_L - V_C = IX_L - IX_C = I(X_L - X_C) = 2\,\text{A}(4\,\Omega - 4\,\Omega) = 0\,\text{V}$

Thus, the voltmeter will read 0 V.

The correct answer is (B) 0V, 2A.

Answer 2

Step 1: Calculate the total impedance (Z) of the circuit.

The circuit consists of a resistor (R), a capacitor ($X_C$), and an inductor ($X_L$) in series. The total impedance (Z) for a series RLC circuit is given by:

$Z = \sqrt{R^2 + (X_L - X_C)^2}$

Given values:

$R = 45\Omega$
$X_L = 4\Omega$
$X_C = 4\Omega$

Substitute the values into the impedance formula:

$Z = \sqrt{(45\Omega)^2 + (4\Omega - 4\Omega)^2}$
$Z = \sqrt{(45\Omega)^2 + (0\Omega)^2}$
$Z = \sqrt{(45\Omega)^2}$
$Z = 45\Omega$

Step 2: Calculate the current (I) in the circuit.

Using Ohm's Law for AC circuits, the current (I) is given by:

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

where V is the RMS voltage of the AC source (90V) and Z is the total impedance (45Ω).

$I = \frac{90V}{45\Omega} = 2A$

The ammeter is in series with the circuit, so it will measure the total current.

Ammeter reading = 2A

Step 3: Calculate the voltage across the capacitor and inductor combination ($V_{LC}$).

The voltmeter is connected across the series combination of the capacitor and inductor. The voltage across the capacitor is $V_C = I \times X_C$ and the voltage across the inductor is $V_L = I \times X_L$.

$V_C = I \times X_C = 2A \times 4\Omega = 8V$

$V_L = I \times X_L = 2A \times 4\Omega = 8V$

In a series RLC circuit, the voltage across the inductor ($V_L$) and capacitor ($V_C$) are 180 degrees out of phase. The net voltage across the LC combination ($V_{LC}$) is the magnitude of the difference between $V_L$ and $V_C$:

$V_{LC} = |V_L - V_C| = |8V - 8V| = 0V$

The voltmeter reads the voltage across the LC combination.

Voltmeter reading = 0V

Step 4: Identify the correct option.

Comparing the calculated readings with the options:

Voltmeter reading = 0V
Ammeter reading = 2A

Option (B) matches these readings: (B) 0V, 2A

Final Answer: The final answer is Option (B) matches these readings: (B) 0V, 2A