JEE Advance - Physics (2020 - Paper 2 Offline - No. 17)
The inductors of two LR circuits are placed next to each other, as shown in the figure. The values of the self-inductance of the inductors, resistors, mutual-inductance and applied voltages are specified in the given circuit. After both the switches are closed simultaneously, the total work done by the batteries against the induced EMF in the inductors by the time the currents reach their steady state values is _________ mJ.
Responder
55
Explicação
Mutual inductance is producing flux in same direction as self-inductance.
$$\therefore$$ $$U = {1 \over 2}{L_1}I_1^2 + {1 \over 2}{L_2}I_2^2 + M{I_1}{I_2}$$
$$ \Rightarrow U = {1 \over 2} \times (10 \times {10^{ - 3}}){1^2} + {1 \over 2} \times (20 \times {10^{ - 3}}) \times {2^2} + (5 \times {10^{ - 3}}) \times 1 \times 2$$
$$ = 55$$ mJ
$$\therefore$$ $$U = {1 \over 2}{L_1}I_1^2 + {1 \over 2}{L_2}I_2^2 + M{I_1}{I_2}$$
$$ \Rightarrow U = {1 \over 2} \times (10 \times {10^{ - 3}}){1^2} + {1 \over 2} \times (20 \times {10^{ - 3}}) \times {2^2} + (5 \times {10^{ - 3}}) \times 1 \times 2$$
$$ = 55$$ mJ