SATHEE: Electro Magnetic Induction And Alternating Currents Question 409

Electro Magnetic Induction And Alternating Currents Question 409

Question: A coil of inductance 300 mH and resistance 2 $Ω$ . is connected to a source of voltage 2 V. The current reaches half of its steady state value in

Options:

A) 0.3s

B) 0.15s

C) 0.1 s

D) 0.05s

Show Answer

Answer:

Correct Answer: C

Solution:

$L = 300 \times 10^{- 3} H$

$R = 1 Ω$

$I = \frac{I_{0}}{2}$ Using $I = I_{0} (1 - e^{- R t / L}),$

we get $\frac{I_{0}}{2} = I_{0} (1 - e^{- R t / L})$

$\Rightarrow \frac{1}{2} = 1 - e^{- R t / L} \Rightarrow e^{- R t / L} = \frac{1}{2}$

$\Rightarrow e^{R t / L} = 2 \Rightarrow \frac{R}{L} t = \log_{e} 2 = 0.693$

$\Rightarrow t = \frac{0.693 \times 300 \times 10^{- 3}}{2} s = 0.1 s$

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Electro Magnetic Induction And Alternating Currents Question 409

Question: A coil of inductance 300 mH and resistance 2 Ω . is connected to a source of voltage 2 V. The current reaches half of its steady state value in

Options:

Answer:

Solution:

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Question: A coil of inductance 300 mH and resistance 2 $Ω$ . is connected to a source of voltage 2 V. The current reaches half of its steady state value in