- 1(current)
Q. No.
An ideal inductor-resistor-battery circuit is switched on at \(t=0~\text{s}\). At time \(t\), the current is \(i=i_0\left(1-e^{\left(-\frac{t}{\tau}\right)}\right)\text{A}\), where \(i_0\) is the steady-state value. The time at which the current becomes \(0.5i_0\) is: [Given \(\text{ln}(2)= 0.693\)]
1. \(6.93 \times 10^3 ~\text{s}\)
2. \(6.93~\text{ms}\)
3. \(69.3~\text{s}\)
4. \(6.93~\text{s}\)
Subtopic: Â LR circuit |
From NCERT
NEET - 2024
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- 1(current)
Q. No.
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