Question

In a given AC circuit consisting of a resistor R and an inductor L and source emf, two voltmeter V₁ & V₂ are connected as shown. If V₂ = 36 volts then inductance of inductor is (Resistance R is √91Ω.)

• A. 0.08 H
• B. 0.8 H
• C. 8 H
• D. 80 H

Answer 1

The Correct Option is A

To find the inductance of the inductor, we can use the given information and apply the concept of AC circuits, where a resistor and an inductor are connected in series.

In this circuit, the voltage drop across the resistor \( V_2 = 36 \) volts and the resistance \( R = \sqrt{91} \, \Omega \).

Since the resistor and inductor are in series, the total voltage across the combination is equal to the sum of the voltages across the resistor and the inductor. According to Ohm's law, the voltage across the resistor can be given as:

V = I \cdot R

Given that V_2 = 36 \, \text{volts}, we can find the current I:

I = \frac{V_2}{R} = \frac{36}{\sqrt{91}}

Now, the inductive reactance X_L is given by:

X_L = I \cdot L = 2 \cdot \pi \cdot f \cdot L

However, the problem does not mention the source frequency explicitly. To find L, we use the relationship between the voltages:

Since V_1 and V_2 are across the inductor and resistor, respectively, in a standard AC series circuit, they relate to each other through the impedances:

\text{Impedance } (Z) = \sqrt{R^2 + (X_L)^2}

The total voltage V\_{\text{source}} = 120 \, \text{volts} (as shown in the image) can be related as:

V\_{\text{source}}^2 = V_1^2 + V_2^2

Substitute the known values:

120^2 = V_1^2 + 36^2

Solve for V_1:

V_1 = \sqrt{120^2 - 36^2} \approx 115.5 \, \text{volts}

Using V_1 = I \cdot X_L, we find:

I \cdot 2 \cdot \pi \cdot f \cdot L = \sqrt{120^2 - 36^2}

X_L = \frac{115.5}{I}

We know current I:

I = \frac{36}{\sqrt{91}}

Inductive reactance leads us to:

L = \frac{115.5}{2 \cdot \pi \cdot f \cdot I}, assuming common frequency for household AC (f = 60 \, \text{Hz}).

Perform the calculations to find L:

Thus, the inductance \( L = 0.08 \, \text{H} \).