SATHEE: Electromagnetic Induction and Alternating Current 4 Question 2

Electromagnetic Induction and Alternating Current 4 Question 2

####2. A coil of self inductance $10 mH$ and resistance $0.1 Ω$ is connected through a switch to a battery of internal resistance $0.9 Ω$ . After the switch is closed, the time taken for the current to attain $80$ of the saturation value is [Take, $\ln 5 = 1.6$ ]

(Main 2019, 10 April II)

(a) $0.002 s$

(b) $0.324 s$

(d) $0.016 s$

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Answer:

Correct Answer: 2. (d)

Solution:

Key Idea In an $L - R$ circuit, current during charging of inductor is given by

$i = i_{0} (1 - e^{- \frac{R}{L} \cdot t})$

where, $i_{0} =$ saturation current.

In given circuit,

Inductance of circuit is

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

Resistance of circuit is

$R = (R_{s} + r) = 0.1 + 0.9 = 1 Ω$

$R_{S} = 0.1 Ω$

Now, from

$\begin{aligned} Given, & i & = i_{0} (1 - e^{- \frac{R}{L} \cdot t}) \\ \Rightarrow & i & = 80 & i & = \frac{80 i_{0}}{100} = 0.8 i_{0} \end{aligned}$

Substituting the value of $i$ in Eq. (i), we get

$\begin{aligned} 0.8 = 1 - e^{- \frac{R}{L} t} \Rightarrow e^{- \frac{R}{L} t} = 0.2 \Rightarrow e^{\frac{R}{L} t} = 5 & \Rightarrow \ln (e)^{\frac{R}{L} t} = \ln 5 \Rightarrow \frac{R}{L} t = \ln 5 & \Rightarrow t = \frac{L}{R} \cdot \ln (5) = \frac{10 \times 10^{- 3}}{1} \times \ln (5) & = 10 \times 10^{- 3} \times 1.6 & = 1.6 \times 10^{- 2} s = 0.016 s \end{aligned}$

Electromagnetic Induction and Alternating Current 4 Question 2

Answer:

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Electromagnetic Induction and Alternating Current 4 Question 2

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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