Question

A coil has resistance $30$ ohm and inductive reactance $20$ ohm at $50 \,Hz$ frequency. If an ac source, of $200$ volt, $100\, Hz$, 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, $f =50 Hz$
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 $100 Hz$,
$X _{ L }^{\prime}=\omega L =2 \pi \cdot 100 \cdot L$
$\Rightarrow X _{ L }^{\prime}=2 \pi \cdot(50 \times 2) L$
$\Rightarrow X _{ L }^{\prime}=20 \times 2 \Omega(\text { from (1) })$
$\Rightarrow X _{ L }^{\prime}=40 \Omega$
Impedance, $Z =\sqrt{ X _{ K }^{2}+ R ^{2}}$
$=\sqrt{(40)^{2}+(30)^{2}}$
$Z=50 \Omega$
Current, $I =\frac{ V }{ Z }=\frac{200}{50}=4 A$