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

Correct answer: A. 4.0 ~A If ω=50 × 2 π then ω L=20 Similarly ω^ =100 × 2 π then ω^ L=40 aligned i & =(200)/(Z)= 200 R^2+(ω^ L)^2 & = 200 (30)^2+(40)^2 =4 ~A aligned

NEET 2011

Exam: NEET
Year: 2011
Subject: Physics
Chapter: Alternating Current

NEET 2011 · Physics · Alternating Current

A coil has resistance 30 and inductive reactance 20 at 50 ~Hz...

A coil has resistance

30 \Omega

and inductive reactance

20 \Omega

50 \mathrm{~Hz}

frequency. If an

\mathrm{AC}

source of

200 \mathrm{~V}, 100 \mathrm{~Hz}

, is connected across the coil, the current in the coil will be

Options

A $4.0 \mathrm{~A}$
B $8.0 \mathrm{~A}$
C $\frac{20}{\sqrt{13}} \mathrm{~A}$
D $2.0 \mathrm{~A}$

Correct answer

A. $4.0 \mathrm{~A}$

Step-by-step solution

\omega=50 \times 2 \pi

then

\omega L=20 \Omega

Similarly

\omega^{\prime}=100 \times 2 \pi

then

\omega^{\prime} L=40 \Omega

\begin{aligned}i & =\frac{200}{Z}=\frac{200}{\sqrt{R^2+\left(\omega^{\prime} L\right)^2}}\\ & =\frac{200}{\sqrt{(30)^2+(40)^2}}=4 \mathrm{~A} \end{aligned}

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