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
In an ideal transformer the voltages are in the ratio of the turns. The bulb is connected to the secondary, so first find the secondary voltage and then use Ohm’s law for the bulb.
\( \displaystyle \frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow V_s = 240\,\mathrm{V}\times\frac{100}{4000} = 6\,\mathrm{V} \)
\( \displaystyle I_s = \frac{V_s}{R} = \frac{6\,\mathrm{V}}{1.2\,\Omega} = 5\,\mathrm{A} \)