Home
Class 12
PHYSICS
When a voltage of 120V is impressed acro...

When a voltage of 120V is impressed across the primary of a transformer , the current in the primary is 1.85mA . Find the voltage across the secondary , when it delivers 150mA . The transformer has an efficiency of 95%

Text Solution

Verified by Experts

Data supplied ,
`V_P = 120 V, I_P= 1.85 A , I_S = 150 mA = 150 xx 10^(-3)A `
`eta = (V_S I_S)/(V_P I_P)`
`V_S = (eta V_P I_P)/(I_S) = (0.95 xx 120 xx 1.85)/(150 xx 10^(-3) ) = 1406 V`
Promotional Banner

Topper's Solved these Questions

  • ALTERNATING CURRENT

    NEW JOYTHI PUBLICATION|Exercise SOLUTIONS TO EXERCISES FROM NCERT TEST|24 Videos
  • ALTERNATING CURRENT

    NEW JOYTHI PUBLICATION|Exercise PRACTICE PROBLEMS FOR SELF ASSESSMENT|11 Videos
  • ATOMS

    NEW JOYTHI PUBLICATION|Exercise COMPETITIVE EXAM CORNER|5 Videos

Similar Questions

Explore conceptually related problems

An a.c voltage of 300V applied to the primary of a transformer & voltage of 3000V is obtained from the secondary coil. Calculate the ratio of circuit in the primary & secondary coils.

The voltage across a wire is (100pm5)V and the current passing through it is (10pm0.2) A. Find the resistance of the wire.

The 300 turn primary of a transformer has resistnace 0.82Omega and the resistance of its secondary of 1200 turns is 6.2Omega . Find the voltage across the primary if the power output from the secondary at 1600 V is 32 kW. Calculate the power losses in both coils when the transformer efficiency is 80% .

A transformer has 50 turns in the primary and 100 in the secondary. If the primary is connected to a 220 V DC supply, what will be the voltage across the secondary?

A transformer having efficiency of 90% is working on 200 V and 3 kW power supply.If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively are

When 115 V is applied across a wire of length 10m and radius 0.3 mm, the current density is 1.4 xx 10^4 A m^2 . The resistivity of the wire is: