Home
Class 12
PHYSICS
In the circuit shown the cell is ideal. ...

In the circuit shown the cell is ideal. The codil has an inductance of 2H and will blow when the current through it reaches 5a. The switch is closed at t=0. Find the time (in second)n when fuse will blow.

Text Solution

Verified by Experts

The correct Answer is:
5
Promotional Banner

Similar Questions

Explore conceptually related problems

In the circuit shows Fig the cell is ideal. The coil has an inductance of 4 H and zero resistance. F is a fuse of zero resistance and will blow when the current through it reaches 5 H . The switch is closed at t= 0 . The fuse will blow

The cell in the circuit shows in Fig is ideal. The coil has an inductance of 4 mH and a resistance of 2 mOmega . The switch is closed at t = 0 . The amount of energy stored in the inductor at t = 2 s is (take e = 3)

In the circuit shown in figure the switch S is closed at t = 0. A long time after closing the switch

In the circuit shown, the switch 'S' is closed at t = 0 . Then the current through the battery steady state reached is

In the circuit shown in figure, circuit is closed at time t=0 . At time t=ln(2) second

In the circuit shown, the switch is closed at t=0 , the currents I_1, I_2 and I_3 are

In the circuit shown, switch S is closed at time t = 0 . Find the current through the inductor as a function of time t .

Switch is closed at t = 0 then the current in the circuit at t = L/2R is

In the circuit diagram shown, initially there is no energy in the inductor and the capacitor, The switch is closed at t = 0 . Find the current I as a function of time if R=sqrt(L//C)

Passage-III In the circuit shown, the battery is ideal with emf V. The capacitor is initially uncharged. The switch is closed at time t = 0 The current in the branch BE at t=oo is