- 1(current)
Q. No.In an AC sub-circuit, the resistance \(R=0.2~\Omega. \) At a certain instant \((V_{A}-V_{B})=0.5~\text{V}, \) \(I=0.5~\text{A}, \) and current is increasing at the rate of \(\left(\frac{d i}{d t}\right)=8~\text{A/s} \). The inductance of the coil is:
1. \(0.01~\text{H}\)
2. \(0.02~\text{H}\)
3. \(0.05~\text{H}\)
4. \(0.5~\text{H}\)
1. \(0.01~\text{H}\)
2. \(0.02~\text{H}\)
3. \(0.05~\text{H}\)
4. \(0.5~\text{H}\)
Subtopic: Â LR circuit |
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In a part of the circuit, as shown, it is given that the current is decreasing at a rate of \(1~\text{A/s}.\) Then \(V_A-V_B\) is equal to:
1. \(18~\text{V}\)
2. \(-18~\text{V}\)
3. \(9~\text{V}\)
4. \(-9~\text{V}\)
1. \(18~\text{V}\)
2. \(-18~\text{V}\)
3. \(9~\text{V}\)
4. \(-9~\text{V}\)
Subtopic: Â LR circuit |
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From NCERT
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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 |
 61%
From NCERT
NEET - 2024
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An inductor \((L)\) and a resistor \((R)\) are connected in series across a battery of emf \(E,\) and the circuit is switched on. The current rises steadily. The rate of increase of the current \(\left(\text{i.e.,}\dfrac {di} {dt}\right),\) when the voltage drops across the resistor is \(\dfrac{E}{2}\), is given by: \(\dfrac {di} {dt}\) =
| 1. | \(\dfrac{E}{L}\) | 2. | \(\dfrac{E}{2L}\) |
| 3. | \(\dfrac{2E}{L}\) | 4. | \(\dfrac{E}{L}e^{-1}\) |
Subtopic: Â LR circuit |
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From NCERT
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The current in the branch of a circuit shown below increases at a rate of \(3\) A/s. At the instant when the current in the wire is \(2\) A, the potential drop from \(a\) to \(b\) is:
1. \(26~\text{V}\)
2. \(14~\text{V}\)
3. \(16~\text{V}\)
4. \(6~\text{V}\)
Subtopic: Â LR circuit |
 68%
From NCERT
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- 1(current)
Q. No.
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