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  • An inductor (L = 0.03 H) and a resistor are connected in series to a battery of 15 V emf

An inductor (L = 0.03 H) and a resistor (R=0.15\Omega )are connected in series to a battery of 15 V emf in a circuit shown below. The key 1 K has been kept closed for a long time. Then, at t = 0, K_{t} is opened and key 2 K is closed simultaneously. At t = 1 m/s the current in the circuit will be \mathrm{(e^{5}\cong 150)} :

 

Option: 1

100 mA


Option: 2

67 mA


Option: 3

6.7 mA


Option: 4

0.67 mA


Answers (1)

best_answer

When switch K_{1} closed for long time current through inductor, \mathrm{I=\frac{E}{R}-\frac{15}{0.15 \times 10^3}=0.1 \mathrm{~A}}

\mathrm{\text { When } \mathrm{K}_1 \text { opened and } \mathrm{K}_2 \text { is closed, then }}

\mathrm{\begin{aligned} & i-I e^{-1 / t} \\ & t=1 \mathrm{~ms}=1 \times 10^{-3} \mathrm{~s} \end{aligned}}

\mathrm{\begin{aligned} & \therefore \quad \tau_L=\frac{L}{R}=\frac{0.03}{0.15 \times 10^3}=\frac{10^{-3}}{5} \\ & \left.\therefore \quad i=0.1 e^{\left(\frac{-1 \times 10^{-1}}{10^{-1}}\right.}\right)=0.1 e^{-5} \\ & \quad i=\frac{0.1}{e^5}=\frac{0.1}{150}=\frac{1}{15} \times 10^{-2} \times 10^3 \mathrm{~mA} \\ & i=0.67 \mathrm{~mA} \end{aligned}}

 

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Anam Khan

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