Question 137: AC fundamentals

Question

A coil of negligible resistance is connected in series with $$90 \Omega$$ resistor across $$120 \mathrm{~V}, 60 \mathrm{~Hz}$$ supply. A voltmeter reads $$36 \mathrm{~V}$$ across resistance. Inductance of the coil is :

Accepted Answer

To find the inductance of the coil, we need to analyze the given circuit and use the information provided. The circuit consists of a resistor and an inductor in series, connected to an AC supply. Here are the given values:
1. Resistance, $$R = 90 \Omega$$
2. Supply voltage, $$V_{	ext{total}} = 120 \mathrm{~V}$$
3. Frequency, $$f = 60 \mathrm{~Hz}$$
4. Voltage across the resistor, $$V_R = 36 \mathrm{~V}$$
First, we calculate the current through the resistor (which is the same as the current through the inductor, since they are in series) using Ohm's law:
$$I = \frac{V_R}{R} = \frac{36}{90} = 0.4 \mathrm{~A}$$
Next, we find the total impedance $$Z$$ of the series combination from the total supply voltage:
$$V_{	ext{total}} = I \cdot Z$$
$$Z = \frac{V_{	ext{total}}}{I} = \frac{120}{0.4} = 300 \Omega$$
We know that the total impedance in a series circuit consisting of a resistor and an inductor is given by:
$$Z = \sqrt{R^2 + (X_L)^2}$$
where $$X_L$$ is the inductive reactance. Rearrange this to solve for $$X_L$$:
$$X_L = \sqrt{Z^2 - R^2} = \sqrt{(300)^2 - (90)^2} = \sqrt{90000 - 8100} = \sqrt{81900} \approx 286 \Omega$$
Now, we use the inductive reactance formula to find the inductance $$L$$:
$$X_L = 2\pi f L$$
$$L = \frac{X_L}{2\pi f} = \frac{286}{2 \pi \cdot 60} \approx \frac{286}{376.99} \approx 0.76 \mathrm{~H}$$
Thus, the inductance of the coil is:
Option B: 0.76 H

Answer

0.91 H

Answer

0.76 H

Answer

2.86 H

Answer

0.286 H

AC fundamentals

A coil of negligible resistance is connected in series with $90 \Omega resistor across 120 \mathrm{~V}, 60 \mathrm{~Hz} supply. A voltmeter reads 36 \mathrm{~V}$ across resistance. Inductance of the coil is :

Choose the correct answer: