A transformer has 1500 turns in the primary coil and 1125 turns in the secondary coil. If the voltage in the primary coil is 200 V, then the voltage in the secondary coil is

100 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

150V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

200V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

250V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 150V
$\frac{N_{P}}{N_{S}} = \frac{1500}{1125} = \frac{200}{V_{S}}$
$V_{S} = \frac{2 \times 1125}{15} = \frac{2 \times 225}{3}$
$= \frac{450}{3}$
$V_{s} = 150 V$

Suggest Corrections

Similar questions

A transformer has $1500$ turns in the primary coil and $1125$ turns in the secondary coil. If the voltage in the primary coil is $200 V$ , then the voltage in the secondary coil is

Q. A transformer has

200

turns in primary coil and

600

turns in secondary coil. If a

220 V

D C

is applied across the primary coil, what will be the voltage across the secondary coil.

Q. The primary coil of a transformer has 100 turns and the secondary coil has 200 turns. The primary coil is connected to a size AA battery that supplies a constant voltage of 1.5 volts. What is the measured voltage across the secondary coil?

Q. The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primary coil is connected to a size AA battery that supplies a constant voltage of 1.5 volts.The voltage in the secondary coil will be volts.

Q. In a transformer, there are

10, 000

turns in primary coil and

25, 000

turns in secondary coil. An alternating voltage,

E = 50 sin (π t)

is applied across the primary coil. Find the peak voltage across the secondary coil, in ideal conditions.

Join BYJU'S Learning Program

A transformer has 1500 turns in the primary coil and 1125 turns in the secondary coil. If the voltage in the primary coil is 200 V, then the voltage in the secondary coil is

The correct option is B 150VNPNS=15001125=200VS VS=2×112515=2×2253 =4503 Vs=150V

The correct option is B 150V
$\frac{N_{P}}{N_{S}} = \frac{1500}{1125} = \frac{200}{V_{S}}$
$V_{S} = \frac{2 \times 1125}{15} = \frac{2 \times 225}{3}$
$= \frac{450}{3}$
$V_{s} = 150 V$