Advertisements
Advertisements
Question
A coil of self-inductance 2.5H and resistance 20Ω is connected to a battery of emf 120V having the internal resistance of 5 n. Find:
1) The time constant of the circuit.
2) The current in the circuit in steady state
Advertisements
Solution
Given
l = 2.5h
R = 20Ω
E = 120V
r = 5Ω
1) Tme constant , `t = L/R = 2.5/20 = 1/8 = 0.125 s`
2) Current, `I = E/(R + r) =- 120/25 = 4.8 A`
APPEARS IN
RELATED QUESTIONS
If the radius of a sphere is doubled without changing the charge on it, then electric flux originating from the sphere is ______.
An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10−3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.
Explain the phenomenon of mutual induction.
Find the mutual inductance between the circular coil and the loop shown in figure.
In mutual induction, the main current remains same because ____________.
A current I = 10 sin (50 π t) ampere is passed in the first coil which induces a maximum e.m.f. of 5 π volt in the second coil. The mutual inductance between the coils is ______.
Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon ______.
What does the coupling factor \[K\] quantify?
What is the total flux linkage of secondary solenoid \[S_2\] ?
