Question

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

• A. 2.0 A
• B. 4.0 A
• C. 8.0 A
• D. $\frac{20}{\sqrt{13}}A$

Answer 1

Answer: (B)

Given,
Resistance, $R=30\Omega$
Inductive Reactance, $X_L=20\Omega$
Frequency, $f=50Hz$
We know,
$X_L=\omega L=2\pi fL$
$20=X_L=2\pi \cdot 50\cdot L$.......(1)
When frequency of ac source is changed to $100Hz$,
$X_L^{\prime }=\omega L=2\pi \cdot 100\cdot L$
$\Rightarrow X_L^{\prime }=2\pi \cdot (50\times 2)L$
$\Rightarrow X_L^{\prime }=20\times 2\Omega (\text{ from (1) })$
$\Rightarrow X_L^{\prime }=40\Omega$
Impedance, $Z=\sqrt{X_K^2+R^2}$
$=\sqrt{(40{)}^2+(30{)}^2}$
$Z=50\Omega$
Current, $I=\frac{V}{Z}=\frac{200}{50}=4A$