I am quite new at circuit analysis and I have a question about a RC circuit, with the resistor and the capacitor connected in series. I want to find the energy supplied by the ideal voltage source given that the capacitor is not charged, until it's fully charged. However, I am a bit lost because of the resistor, which I am sure that will also consume some energy. Any advice on how to proceed?
Answer
The energy balance can be considered by noting that the work done in moving a charge $\delta q$ across a potential difference $V$ is $V \delta q$ and an integration can be done to find the total amount of work done $\int V dq$ whilst moving a finite amount of charge.
This integration is equivalent to finding the area under a potential difference against charge graph.
Applying KVL to a series battery, resistor and capacitor circuit gives the relation between the emf of the battery $\mathcal E$, the potential difference across the resistor $v_R$ and the potential difference across the capacitor $v_C$
$\mathcal E - v_R - v_C = 0$
When the capacitor is fully charged $v_R = 0$ and $v_C = \mathcal E$.
The relationship between the potential difference $v_C$ across a capacitor of capacitance $C$ and the charge stored $q$ is $v = \frac 1 C q$
Assume that the capacitor is originally uncharged and that the final charge stored on it is $Q$ and then the potential difference across the capacitor is $\mathcal E$.
Graph 1 shows the variation of potential difference with charge stored.
The work done is the area under the graph and equal to the energy stored in the capacitor $\frac 1 2 \mathcal E Q$.
Since $v_R = \mathcal E - v_C$ graph 2 shows the variation of voltage across the resistor and the charge which has flowed through the resistor.
The area under this graph is the work done by the battery in driving charge through the resistor which is equal to the energy dissipated in the resistor as heat.
That again is $\frac 1 2 \mathcal E Q$.
In charging the capacitor the battery has driving a charge $Q$ around the circuit and has done work equal to $\mathcal E Q$ and this has come from a chemical reaction within the battery.
This is graph 3.
So using KVL which comes from the law of conservation of energy, the definition of potential difference and the defining equation for capacitance we have
Electrical energy supplied by the battery $(\mathcal C \mathcal E^2)$
= energy stored in the capacitor $(\frac 12 C \mathcal E^2)$ + energy dissipated as heat in the resistor $(\frac 12 C \mathcal E^2)$.
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