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Figure shows a part of a bigger circuit. The capacity of the capacitor is `6 mF` and is desaesing at the constate rate `0.5 mF s^(-1)`. The potential difference across the capacitor at the shows moment is changing as follows:
`(dV)/(dt) = 2 V s^(-1)`, `(d^(2)V)/(dt^(2)) = (1)/(2) V S^(-2)`
The current in the `4 Omega` resistor is decreasing at the rate of `1 mA s^(-1)`. What is the potential difference (in `mV`) across the inductor at this moment?

Text Solution

Verified by Experts

The correct Answer is:
`(4)`
`q = CV,I = (dq)/(dt) = C(d nu)/(dt) = V(dC)/(dt), I_(L) = I_(0) - I`

`(dI_(L))/(dt) = (dI_(0))/(dt) - [C(d^(2)V)/(dt^(2)) + (dV)/(dt)(dC)/(dt) + (dV)/(dt)(dC)/(dt) + V (d^(2)C)/(dt^(2))]`
`= - 1 xx 10^(-3) - [6 xx 10^(-3) xx (1)/(2) + 2(-0.5 xx 10^(-3)) + 2(-0.5 xx 10^(-3)) + 10]`
`= - 2 xx 10^(-3) A//s`
`V_(L) = L(dI_(L))/(dt) = 2 xx 2 xx 10^(-3) = 4xx10^(-3) V = 4 mV`
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