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In the circuit shown in figure, circuit ...

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

A

rate of energy supplied by the battery is 16 J/s

B

rate of heat dissipated across resistance is 8J/s

C

rate of heat dissipated across resistance is 16 J/s

D

`V_a-V_b=4V`

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
`tau_L=L/R=2/2=1s`
`t 1/2=(ln 2) tau_L=(ln 2) s`
Hence the given time is half time.
`:. i=i_0/2=(8//2)/2=2A`
Rate of energy supplied by battery
`=Ei=8xx2=16J//s`
`P_R=i^2R=(2)^2(2)=8J//s`
`V_a-V_b=E-iR=9-2xx2=4V`
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