Electromagnetic Induction

JEE Main level, numerical answer, medium

102 of 153

Question 102 of 153

In the given circuit below inductance values of L1L_1, L2L_2 and L3L_3 are same. The magnetic energy stored in the entire circuit is (Ut)(U_t) and that stored in the L2L_2 inductor is (Ul)(U_l). Ut/UlU_t / U_l is __________.
(Ignore the mutual inductance if any)

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Answer: 6

Let the inductance of each inductor be LL, so L1=L2=L3=LL_1 = L_2 = L_3 = L.

Let the total current flowing through the circuit be II. This current flows entirely through L1L_1.

Since L2L_2 and L3L_3 are connected in parallel and have equal inductances, the current II divides equally between them. Therefore, the current through L2L_2 is I2\dfrac{I}{2} and the current through L3L_3 is I2\dfrac{I}{2}.

The magnetic energy stored in an inductor is given by U=12Li2U = \dfrac{1}{2} L i^2.

The energy stored in the inductor L2L_2 is:
Ul=12L(I2)2=18LI2U_l = \dfrac{1}{2} L \left(\dfrac{I}{2}\right)^2 = \dfrac{1}{8} L I^2

The equivalent inductance of the entire circuit is:
Leq=L1+L2L3L2+L3=L+L⋅LL+L=L+L2=3L2L_{eq} = L_1 + \dfrac{L_2 L_3}{L_2 + L_3} = L + \dfrac{L \cdot L}{L + L} = L + \dfrac{L}{2} = \dfrac{3L}{2}

The total magnetic energy stored in the entire circuit is:
Ut=12LeqI2=12(3L2)I2=34LI2U_t = \dfrac{1}{2} L_{eq} I^2 = \dfrac{1}{2} \left(\dfrac{3L}{2}\right) I^2 = \dfrac{3}{4} L I^2

The ratio of the total energy to the energy stored in L2L_2 is:
UtUl=34LI218LI2=34×8=6\dfrac{U_t}{U_l} = \dfrac{\dfrac{3}{4} L I^2}{\dfrac{1}{8} L I^2} = \dfrac{3}{4} \times 8 = 6

Answer: 66

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