An inductor of inductance L= 400mH and resistors R1=2Ω and R2=2Ω are connected to a battery of EMF 12V . The internal resistance of the battery is negligible. The switch S is closed at t=0. The potential drop across L as a function of time is?

The voltage across the inductor in a circuit is directly proportional to the rate of flow of current through the inductor.

For this firstly calculate the current flowing through the branch containing L and R2. Then differentiate current I with respect to time t.

Given: L = 400mH

R1 = 2Ω

R2 = 2Ω

E = 12V

In the branch containing L and R2, I = E/R₂ 1−e^−R2t/L

Differentiating I with respect to t:

dI/dt = ER2e^−R2t/L. R2/L

= E/Le^−R2t/L

Therefore, VL = L dI/dt = Ee^−R2t/L

= 12^e−5tV

Then the potential drop across L as a function of time is 12^e−5tV.

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An inductor of inductance L= 400mH and resistors R1=2\(\Omega\) and R2=2\(\Omega\) are connected to a battery of EMF 12V . The internal resistance of the battery is negligible. The switch S is closed at t=0. The potential drop across L as a function of tim

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An inductor of inductance L= 400mH and resistors R1=2\(\Omega\) and R2=2\(\Omega\) are connected to a battery of EMF 12V . The internal resistance of the battery is negligible. The switch S is closed at t=0. The potential drop across L as a function of tim

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2.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

3.Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

4.A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

5.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

6. If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

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Comments

SUBSCRIBE TO OUR NEWS LETTER

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