A transformer has 1000 turns in the primary and 2000 turns in its secondary winding. The primary voltage is 200 V and the load across the secondary is 100 Ω. If the transformer is assumed to be ideal, then the value of current in the primary is

8.0 A

The correct answer is Option (4) → 8.0 A

Primary turns: $N_p = 1000$, Secondary turns: $N_s = 2000$

Primary voltage: $V_p = 200 \, V$

Secondary voltage: $V_s = \frac{N_s}{N_p} V_p = \frac{2000}{1000} \times 200 = 400 \, V$

Load resistance: $R = 100 \, \Omega$

Secondary current: $I_s = \frac{V_s}{R} = \frac{400}{100} = 4 \, A$

For ideal transformer: $V_p I_p = V_s I_s$

$I_p = \frac{V_s I_s}{V_p} = \frac{400 \times 4}{200} = 8 \, A$

Answer: $ \; 8 \, A $