JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020
Answers
Answer:
determination of GHSS physics question number 2020 given circuit of both regions of batteries against
The inductor of two LR circuits are placed next to each other , as shown in figure. the value of the self inductance of the inductors, resistors, mutual inductance and applied voltages are specified in the given circuit. after both switches are closed simultaneously, the total work done by the batteries against the induced EMF in the inductors by the time the current reach their steady state value is...
solution : here mutual inductance will produce flux in the same direction as self inductance.
total energy in a coupled circuit for this situation, U = 1/2 L₁i₁² + 1/2 L₂i₂² + Mi₁i₂
= 1/2 × (10 × 10^-3) × 1² + 1/2 × (20 × 10^-3) × 2² + (5 × 10^-3) × 1 × 2
= 5 × 10^-3 + 40 × 10^-3 + 10 × 10^-3
= 55 × 10^-3 J
= 55 mJ
from work energy theorem,
work done by the batteries against induced emf in the inductors = U = 55 mJ