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:
Answer & explanation
Correct answer: option 4
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]
$I_s=\frac{6}{1.2}=5A$