Physics-
General
Easy
Question
An inductor (L = 100 mH), a resistor (R = 100 ) and a battery (E = 100 V) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit, 1 ms after the short circuit is:
- 1 A
- 1/e A
- e A
- 0.1 A
The correct answer is: 1/e A
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One conducting u tube can slide inside another as shown in figure, maintaining electrical contacts between the tubes. The magnetic field B is perpendicular to the plane of the figure. If each tube moves towards the other at a constant speed v, then the emf induced in the circuit in terms of B, and v, where is the width of each tube, will be
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Figure shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance 1 /m. Position of the conducting rod at t = 0 is shown. A time t dependent magnetic field B = 2t Tesla is switched on at t = 0
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Figure shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance 1 /m. Position of the conducting rod at t = 0 is shown. A time t dependent magnetic field B = 2t Tesla is switched on at t = 0
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Figure shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance 1 /m. Position of the conducting rod at t = 0 is shown. A time t dependent magnetic field B = 2t Tesla is switched on at t = 0
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An inductor having self-inductance L with its coil resistance R is connected across a battery of emf . When the circuit is in steady state at t = 0 an iron rod is inserted into the inductor due to which its inductance becomes n L (n > 1)
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An inductor having self-inductance L with its coil resistance R is connected across a battery of emf . When the circuit is in steady state at t = 0 an iron rod is inserted into the inductor due to which its inductance becomes n L (n > 1)
When again circuit is in steady state, the current in it is :
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An inductor having self-inductance L with its coil resistance R is connected across a battery of emf . When the circuit is in steady state at t = 0 an iron rod is inserted into the inductor due to which its inductance becomes n L (n > 1)
After insertion of the rod, current in the circuit:
An inductor having self-inductance L with its coil resistance R is connected across a battery of emf . When the circuit is in steady state at t = 0 an iron rod is inserted into the inductor due to which its inductance becomes n L (n > 1)
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An inductor having self-inductance L with its coil resistance R is connected across a battery of emf . When the circuit is in steady state at t = 0 an iron rod is inserted into the inductor due to which its inductance becomes nL (n > 1)
After insertion of rod which of the following quantities will change with time?
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3) Rate of heat produced in coil
An inductor having self-inductance L with its coil resistance R is connected across a battery of emf . When the circuit is in steady state at t = 0 an iron rod is inserted into the inductor due to which its inductance becomes nL (n > 1)
After insertion of rod which of the following quantities will change with time?
1) Potential difference across terminals A and B.
2) Inductance.
3) Rate of heat produced in coil
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Statement-2 : In the situation of statement-1, the direction in which the rod will slide is that which tends to maintain constant flux through the loop. Providing a larger loop area counteracts the decrease in magnetic flux. So the rod moves to the right independent of the fact that the direction of magnetic field is into the page or out of the page.
Statement-1 : Consider the arrangement shown below. A smooth conducting rod, CD, is lying on a smooth U-shaped conducting wire making good electrical contact with it. The U-shape conducting wire is fixed and lies in horizontal plane. There is a uniform and constant magnetic field B in vertical direction (perpendicular to plane of page in figure). If the magnetic field strength is decreased, the rod moves towards right.
Statement-2 : In the situation of statement-1, the direction in which the rod will slide is that which tends to maintain constant flux through the loop. Providing a larger loop area counteracts the decrease in magnetic flux. So the rod moves to the right independent of the fact that the direction of magnetic field is into the page or out of the page.
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Statement-1 : A resistance R is connected between the two ends of the parallel smooth conducting rails. A conducting rod lies on these fixed horizontal rails and a uniform constant magnetic field B exists perpendicular to the plane of the rails as shown in the figure. If the rod is given a velocity v and released as shown in figure, it will stop after some time. The total work done by magnetic field is negative.
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In the circuit diagram shown
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physics-General