Question

A coil has resistance 30 ohm and inductive reactance 20 ohm at 50Hz frequency. If an ac source, of 200volt, 100Hz, is connected across the coil, the current in the coil will be

Answer 1

Step 1:Given information:

Resistance, $R=30\Omega$

Inductive Reactance,$X_L=20\Omega$

Frequency, $f=50Hz$

Step 2:

On the basis of the given information:

As we know,

$$\begin{gathered} X_L=\omega L=2\mathrm{\pi fL} \\ 20={\mathrm{X}}_{\mathrm{L}}=2\mathrm{\pi }.50.\mathrm{L} \end{gathered}$$

Now, when the frequency of the AC source is changed to $100Hz$,

$$\begin{gathered} X{\text{'}}_L=\omega \text{'}L=2\mathrm{\pi }.100.\mathrm{L} \\ \mathrm{X}{\text{'}}_{\mathrm{L}}=2\mathrm{\pi }.(50\times 2)\mathrm{L} \\ \mathrm{X}{\text{'}}_{\mathrm{L}}=20\times 2\mathrm{\Omega } \\ \mathrm{X}{\text{'}}_{\mathrm{L}}=40\mathrm{\Omega } \end{gathered}$$

Now Impedance, $Z=\sqrt{X{{\text{'}}^2}_K+R^2}$

$Z=\sqrt{{\left(40\right)}^2+{\left(30\right)}^2}$

$Z=50\Omega$

And current,

$I=\frac{V}{Z}=\frac{200}{50}$

Therefore, $I=4\Omega$