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AIIMS 201925 May
Exam
AIIMS
Year
2019
Date
25 May
Subject
Physics
Chapter
Alternating Current

AIIMS 2019 · 25 May · Physics · Alternating Current

In a transformer number of turns in primary circuit is 500 and in...

In a transformer number of turns in primary circuit is 500 and in secondary circuit number of turns is 10 . The load resistance is 10Ω10 \Omega and voltage of secondary coil is 50 V50 \mathrm{~V}. Then find the current in primary circuit.

Options

  1. A0.2 A0.2 \mathrm{~A}
  2. B0.3 A0.3 \mathrm{~A}
  3. C0.4 A0.4 \mathrm{~A}
  4. D0.1 A0.1 \mathrm{~A}

Correct answer

D. 0.1 A0.1 \mathrm{~A}

Step-by-step solution

Given, N1=500,N2=10,RL=10ΩV2=50 VN_1=500, N_2=10, R_L=10 \Omega V_2=50 \mathrm{~V} Current in the secondary circuit is i2=V2RL=5010=5 Ai1i2=N2N1⇒i15=10500⇒i1=0.1 A\begin{aligned} & i_2=\frac{V_2}{R_L}=\frac{50}{10}=5 \mathrm{~A} \\ & \frac{i_1}{i_2}=\frac{N_2}{N_1} \Rightarrow \frac{i_1}{5}=\frac{10}{500} \Rightarrow i_1=0.1 \mathrm{~A}\end{aligned}
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