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In the figure shown q is in coulomb and ...

In the figure shown `q` is in coulomb and t in second. At time `t=1 s`

A

`V_a-V_b=4V`

B

`V_b-V_c=1V`

C

`V_c-V_d=16V`

D

`V_a-V_b=20V`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
`q=2t^2`
`i=(dq)/(dt)=4t`
`(di)/(dt)=4A//s`s
At `t=1s, q=2C,i=4A`
and `(di)/(dt)=4A//s`
`V_a-V_b=L(di)/(dt)=1xx4=4V`
`V_b-V_c=q/C=2/2=1V`
`V_C-V_d=ir=4xx4=16V`
`V_a-V_d` is summation of above three i.e. `21V`.
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