Discussion Forum

1)

An inductor (L=0.03H) and a resistor (R= 0.15kΩ) are connected in series to a battery of 15V EMF in a circuit shown below. The key K₁  has been kept closed for a long time. Then at t=0, K₁   is opened and key K₂ is closed simultaneously. At t=1ms, the current in the circuit will be (e⁵  =150)

Answer:

Option D

Explanation:

Central Idea:    After long time inductor behave as short -circuit

  At t=0 , the inductor behaves as short-circuited . The current

              $I_{0}=\frac{E_{0}}{R}=\frac{15V}{0.15kΩ}=100mA$

As K₂   is closed current through the inductor starts decay , which is given at any time t as

     $I=I_{0}e^{\frac{-tR}{L}}= (100mA)e^{\frac{-t\times 15000}{3}}$

At t=1ms

      $I=(100mA)e^{-\frac{1\times 10^{-3}\times15\times 10^{3}}{3}}$

                 $I=(100mA){e^{-5}}=0.6737 mA$

or           I= 0.67 mA