Q171 / 180 · COMEDK 2024 (Afternoon Shift) · Physics

A transformer has 400 turns in its primary winding and 800 turns in its secondary winding. The primary voltage is $20 \mathrm{~V}$ and the load in the secondary is 4 ohm. The current in the primary, assuming it to be an ideal transformer, is

  • A
    2 A
  • B
    40 A
  • C
    4 A
  • Correct answer
    D
    20 A

Solution

For an ideal transformer, the ratio of voltages is equal to the ratio of the number of turns:

$\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$

Given $N_p = 400$, $N_s = 800$, and $V_p = 20 \text{ V}$, we have:

$\dfrac{V_s}{20} = \dfrac{800}{400} = 2$

$V_s = 20 \times 2 = 40 \text{ V}$

The current in the secondary winding $I_s$ is given by Ohm's law with the load resistance $R_s = 4 \text{ } \Omega$:

$I_s = \dfrac{V_s}{R_s} = \dfrac{40}{4} = 10 \text{ A}$

For an ideal transformer, the power input equals the power output, so $V_p I_p = V_s I_s$:

$20 \times I_p = 40 \times 10$

$20 \times I_p = 400$

$I_p = \dfrac{400}{20} = 20 \text{ A}$

Answer: 20 A

Hence, the correct answer is option: 'D'.

Relevant Questions