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In charging a capacitor to a charge $Q$

enter image description here

Is there an equivalent to this in setting up current through an inductive circuit?

Energy stored in inductor $= 1/2Li^2$

What would be the work done in setting up the current $i$ in the inductor?

(I'm in high school, I have trouble deriving the work done for this one. For the capacitor, since the final voltage across it would be the emf of the battery, I can understand the expression for work done by the battery (as shown in the above image) but for inductor it is a little tricky since its voltage is given by $L*(di/dt)$.)

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In an inductive circuit, the current never stops flowing. It tends towards a maximum value (assuming some amount of resistance, for the circuit to heat up.)

If E is the emf of the battery, L is the inductance, and R is the resistance, the battery is constantly doing work even once the terminal current $\frac{E}{R}$ is achieved, and heat is being dissipated in the resistance.

If you work out the integral $$\int_0^{\infty}Eidt = \int_0^\infty\frac{E^2}{R}(1-e^{-Rt/L})$$ it blows up to infinity

In the case of the capacitive circuit, the battery stops doing work once the capacitor is charged, hence we get a finite expression.

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  • $\begingroup$ Thanks! This question was asked back in 1999 in an entrance test and there was no credible source for answers, the ones on the internet being either Li^2 or 1/2Li^2. $\endgroup$
    – jen
    Commented May 15, 2022 at 14:37
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    $\begingroup$ "It tends towards a maximum value" Or, in the case of a constant-current source, stays constant. $\endgroup$
    – Chemomechanics
    Commented May 15, 2022 at 23:01

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