A transformer has 4000 turns in primary and 100 turns in secondary coils. The primary coil in connected to a 240 V and secondary to a bulb of 1.2 Ω resistance. The current in the bulb is:

5.00 A

The correct answer is Option (4) → 5.00 A

The voltage and current relationship in a transformer and related by following equations -

$\frac{V_p}{V_s}=\frac{N_p}{N_s}$  ...(1)

where,

$V_p$ → Primary voltage

$V_s$ → Secondary voltage

$N_p$ → Number of turns in Primary coil

$N_s$ → Number of turns in Secondary coil

$V_p=240 V$

$N_p=4000$ [given]

$N_s=100$

∴, from eqn. (1)

$V_s=\frac{V_p\,N_s}{N_p}=\frac{240×100}{4000}=6V$

Now,

R, resistance of bulb through which secondary coil is connected = 1.2Ω

$∴I_s=\frac{V_s}{R}$ [By Ohm's law]

$I_s=\frac{6}{1.2}=5A$