An initially uncharged capacitor C is fully charged by a device of constant emf connected in series with a resistor R. (a) Show that the final energy stored in the capacitor is half the energy supplied by the emf device. (b) By direct integration of $$i^2R$$ over the charging time, show that the thermal energy dissipated by the resistor is also half the energy supplied by the emf device.
(a) The charge qq on the
capacitor as a function of time is q(t)=(εC)(1−e−t/RC),q(t)=(εC)(1−e−t/RC), so the charging current is i(t)=dq/dt=(dR)e−t/RCi(t)=dq/dt=(dR)e−t/RC .
The energy supplied by the emf is then
U=∫0∞εidt=ε2R∫0∞e−t/RCdt=Cε2=2UcU=∫0∞εidt=ε2R∫0∞e−t/RCdt=Cε2=2Uc
where Uc=12Cε2Uc=12Cε2 is the energy stored in the capacitor.
(b) By directly integrating i2Ri2R
we obtain
UR=∫0∞i2Rdt=ε2R∫0∞e−2t/RCdt=12Cε2UR=∫0∞i2Rdt=ε2R∫0∞e−2t/RCdt=12Cε2
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