Question 78: AC fundamentals

Question

When a coil is connected across a $$20 \mathrm{~V}$$ dc supply, it draws a current of $$5 \mathrm{~A}$$. When it is connected across $$20 \mathrm{~V}, 50 \mathrm{~Hz}$$ ac supply, it draws a current of $$4 \mathrm{~A}$$. The self inductance of the coil is __________  $$\mathrm{mH}$$. (Take $$\pi=3$$)

Accepted Answer

Let's first determine the resistance of the coil when connected to a DC supply. The current drawn by the coil in a DC circuit can be used to calculate its resistance using Ohm's law:
$$R = \frac{V}{I}$$
Given the DC supply voltage $$V_{DC} = 20 \, \mathrm{V}$$ and the current $$I_{DC} = 5 \, \mathrm{A}$$, the resistance $$R$$ is:
$$R = \frac{20 \, \mathrm{V}}{5 \, \mathrm{A}} = 4 \, \Omega$$
Next, let's use the information given for the AC supply. When connected to an AC supply, the coil's impedance $$Z$$ can be determined using the given current. The total voltage and current in an AC circuit are related to the impedance by the formula:
$$Z = \frac{V}{I}$$
Given the AC supply voltage $$V_{AC} = 20 \, \mathrm{V}$$ and the current $$I_{AC} = 4 \, \mathrm{A}$$, the impedance $$Z$$ is:
$$Z = \frac{20 \, \mathrm{V}}{4 \, \mathrm{A}} = 5 \, \Omega$$
The impedance $$Z$$ of the coil in an AC circuit is composed of both the resistance $$R$$ and the inductive reactance $$X_L$$, related by:
$$Z = \sqrt{R^2 + X_L^2}$$
We already know that $$R = 4 \, \Omega$$. We can now solve for the inductive reactance $$X_L$$:
$$5 = \sqrt{4^2 + X_L^2}$$
Squaring both sides of the equation:
$$25 = 16 + X_L^2$$
Solving for $$X_L$$:
$$X_L^2 = 25 - 16$$
$$X_L^2 = 9$$
$$X_L = \sqrt{9}$$
$$X_L = 3 \, \Omega$$
The inductive reactance $$X_L$$ is also related to the inductance $$L$$ and the angular frequency $$\omega$$ by the formula:
$$X_L = \omega L$$
where $$\omega = 2 \pi f$$. Given the frequency $$f = 50 \, \mathrm{Hz}$$ and using $$\pi = 3$$, we find:
$$\omega = 2 	imes 3 	imes 50$$
$$\omega = 300 \, \mathrm{rad/s}$$
Now we can solve for the inductance $$L$$:
$$X_L = 300 L$$
$$3 = 300 L$$
$$L = \frac{3}{300}$$
$$L = 0.01 \, \mathrm{H}$$
Since $$1 \, \mathrm{H} = 1000 \, \mathrm{mH}$$, the self inductance of the coil is:
$$L = 0.01 \, \mathrm{H} 	imes 1000 \, \mathrm{mH/H} = 10 \, \mathrm{mH}$$
Therefore, the self inductance of the coil is $$10 \, \mathrm{mH}$$.

AC fundamentals

When a coil is connected across a $20 \mathrm{~V} dc supply, it draws a current of 5 \mathrm{~A}. When it is connected across 20 \mathrm{~V}, 50 \mathrm{~Hz} ac supply, it draws a current of 4 \mathrm{~A}. The self inductance of the coil is __________ \mathrm{mH}. (Take \pi=3$)

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