question_answer

In the circuit shown, switch \[{{S}_{2}}\]is open and \[{{S}_{1}}\] is closed since long. Take \[E=20\text{ }V,\text{ }L=8.5\text{ }H\]and\[R=10\,\Omega \]. The rate of change of energy stored in the magnetic field inside the inductor, immediately after \[{{S}_{2}}\] is closed, is

A) Increasing at the rate of\[40\text{ }J/s\]                       

B) Decreasing at the rate of \[40\text{ }J/s\]

C) Increasing at the rate of\[20\text{ }J/s\]                         

D) Decreasing at the rate of \[20\text{ }J/s\]

Correct Answer: B

Solution :

[b] Just before closing \[{{S}_{2}},\] the effective circuit is as shown.

\[i=\frac{E}{R}=\frac{20}{10}=2A\] Immediately, after closing \[{{S}_{2}},\] the current through inductor will remain \[\text{i}=2A.\] The effective circuit has been shown in figure.

Applying KVL to the bigger loop, \[(i+{{i}_{1}})R+{{i}_{1}}(2R)=4E\] \[3{{i}_{1}}R=4E-iR\] \[\Rightarrow \,\,\,\,\,\,\,30{{i}_{1}}=80-iR\] \[[\because \,iR=2\times 10=20]\] \[\Rightarrow \,\,\,\,\,\,\,\,{{i}_{1}}=2A\] Applying KVL to loop containing L and cell of emf 3E. \[3E+L\frac{di}{dt}=2R{{i}_{1}}\] \[60+0.5\frac{di}{dt}=40\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\frac{di}{dt}=-40A/s\] Negative sign says that current through the inductor is decreasing energy stored in magnetic field. \[{{U}_{B}}=\frac{1}{2}L{{i}^{2}}\] \[\frac{d{{U}_{B}}}{dt}=Li\frac{di}{dt}=0.5\times 2\times (-40)=-40J/s\] Negative sign says that energy stored in the inductor is decreasing.