Question

An ideal inductor takes a current of 10A when connected to a 25V, 50Hz AC supply. A pure resistor across the same source takes 12.5A. If the two are connected in series across a $100\sqrt{2}V$, 40Hz supply, the current through the circuit will be.

Answer 1

Hint: As a first step, one could read the given question well and hence note down the given values. Then you could find the value of inductance and also resistance from them. After that you could find the net impedance and then find the current through the circuit mentioned accordingly.


Formula used:
Inductive reactance,
$X_L=2\pi L\times f$
Impedance,
$Z=\sqrt{R^2+{X_L}^2}$
Current,
$I={\displaystyle \frac{\epsilon }{Z}}$


Complete step-by-step solution:
In the question, we are given an inductor that is known to take a current of 10A on connecting it to a 25V, 50Hz AC supply. Connecting a pure resistor across this source draws a current of 12.5A. We are supposed to find the current through the circuit if they are connected in series across a $100\sqrt{2}V$, 40Hz source.
For the inductor alone we have,
$\omega L=2\pi fL={\displaystyle \frac{\epsilon }{I}}$
$\Rightarrow 2\pi L={\displaystyle \frac{\epsilon }{fI}}={\displaystyle \frac{125}{50\times 10}}=0.25$…………………………………. (1)
For the resistor we have,
$R={\displaystyle \frac{\epsilon }{I}}={\displaystyle \frac{125}{12.5}}=10\mathrm{\Omega }$…………………………………… (2)
Now for the series connection of the above two components across a $100\sqrt{2}V$, 40Hz source we have,
The inductive reactance as,
$X_L=2\pi L\times f$
From (1),
$X_L=0.25\times 40=10\mathrm{\Omega }$………………………………….. (3)
Now the net impedance of the circuit could be given by,
$Z=\sqrt{R^2+{X_L}^2}$
From (2) and (3),
$Z=\sqrt{{10}^2+{10}^2}=\sqrt{200}=10\sqrt{2}$
The current could be given by,
$I={\displaystyle \frac{\epsilon }{Z}}={\displaystyle \frac{100\sqrt{2}}{10\sqrt{2}}}$
$\therefore I=10A$
Therefore, we found the current through the circuit under given conditions to be 10A.

Note: Impedance is the opposition offered to the current flowing in a circuit. It can be considered similar to the resistance in a purely resistive circuit. But here when capacitor and inductor are present in the circuit, their effects are also taken into account along with the resistance. It is also measured in ohms.