Question

How much work has the battery done from the time the switch was closed until t = 6.51×10-3 s?

Answer 1

Relevant information: The resistance in the circuit is R = 140 Ω, the inductance is L = 2.40 H, and the battery maintains a voltage of ℰ = 30.0 V. At time t=0 the switch is closed.

So I know I have to integrate the power formula
P=IV =I ℰ where I= \[ (ℰ /R)(1-e ^{-Rt/L}) \]
which result in the integrand
\[\int\limits_{0_{}}^{t_0}ℰ(ℰ /R)(1-e ^{-Rt/L}) dt\]
then simplified I got
\[(ℰ^{2}/R ^{2})(t _{0}R +Le^{-Rt/L} -L) \]

just wondering where I might have gone wrong with this integrand

Answer 2

the equation of your circuit is:
\[\large E = RI + L\frac{{dI}}{{dt}}\]
whose solution, using your initial conditions, is:
\[\large I\left( t \right) = \frac{E}{R}\left[ {1 - \exp \left( { - \frac{R}{L}t} \right)} \right]\]
Now, if we can neglect the dissipation of energy due to Joule effect, then the power of your battery is:
\[\large \int\limits_0^{{t_0}} {EI\left( t \right)dt} = \frac{{{E^2}}}{{{R^2}}}\left[ {R{t_0} + L\exp \left( { - \frac{R}{L}{t_0}} \right) - L} \right]\]
So, your answer is right!

Answer 3

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