Question 430: AC fundamentals

Question

An inductor coil takes current 8A when connected to an 100 V and 50 Hz a.c. source. A pure resistor under the same condition takes current of 10A. If inductor coil and resistor are connected in series to an 100V and 40 Hz a.c. supply, then the current in the series combination of above resistor and inductor is

Accepted Answer

Inductive reactance when 100 V, 50 Hz source is connected is given by
$$\mathrm{X_L=\frac{V}{I}=\frac{100}{8}=12.5\Omega}$$
Resistance $$\mathrm{R=\frac{V}{I}=\frac{100}{10}=10\Omega}$$
When 100V, 40 Hz supply is connected the inductive reactance will get reduced
Since it is proportional to the frequency
$$\begin{aligned}
& 	herefore \quad \frac{\mathrm{X}_{\mathrm{L}}^{\prime}}{}=\frac{40}{50}=\frac{4}{5} \
& 	herefore \quad \mathrm{X}_{\mathrm{L}}^{\prime}=\frac{4}{5} \mathrm{X}_{\mathrm{L}}=\frac{4}{5} 	imes 12.5=10 \Omega\end{aligned}$$
The resistance will remain same. When they are connected in series the impedance will be given by
$$\begin{aligned}
& Z=\sqrt{R^2+X_L^{\prime 2}}=\sqrt{(10)^2+(10)^2}=10 \sqrt{2} A \
& I=\frac{V}{Z}=\frac{100}{10 \sqrt{2}}=5 \sqrt{2} \mathrm{~A}
\end{aligned}$$

Answer

$$\frac{10}{\sqrt3}$$A

Answer

$$\frac{5}{\sqrt2}$$A

Answer

10$$\sqrt2$$A

Answer

5$$\sqrt2$$A

AC fundamentals

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