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Question

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, resistances, mutual-inductance and applied voltages are specified in the given circuit. After both the switches are closed simultaneously, then 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.


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Solution

Step 1; Draw the required diagram:

Step 2: Given data:

Note that from the figure,

The resistanceR1=5ohm,R2=10ohm

The voltageV1=5v,V2=20v

The value of inductances-

L1=10mH=10Γ—10-3HL2=20mH=20Γ—10-3H

The value of mutual inductance M=5mH=5Γ—10-3H

Step 3: Formula used:

The emf of LR circuit can be given as

e1=L1dI1dt+MdI2dt

dW1=e1I1dtdW1=L1I1dI1+MI1dI2Similarly,dW2=L2I2dI2+MI2dI1

Total work dW=∫0WdW1+dW2

Step 4: Calculate the total work:

Thus, using the above formula after integrating we get-

Therefore the total workW=12L1I12+12L2I22+MI1I2

Now substitute the given values in the formula,

Therefore the total work

W=12Γ—10Γ—10-312+12Γ—20Γ—10-3Γ—22+5Γ—10-3Γ—1Γ—2W=5Γ—10-3+10Γ—10-3Γ—4+10Γ—10-3W=55mJ

Therefore the total work is 55.


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