![]() I think it is because $C1$ is the "source of voltage" in the right circuit. Using physical principles (not just rules. C1 C2 A) 3 pts Using and, rank from greatest to least the charges on the A capacitors, Q1, Q2, Q3. ![]() The role of $C1$ is certainly a bit different from the other two in this circuit, but I can't accurately explain why it behaves differently as it relates to the voltage and charge, even though it is connected in series with the other two. The capacitors start out discharged, and then a battery (not shown) is connected between points A and B and all three capacitors become fully charged. Instead, the potential across $C1$ is given via Kirchhoff's law : $V1 -V2 -V3 = 0$. However, the potential across $C1$ cannot be calculated this way, nor is the charge equal to the charge across $C2$ and $C3$. In series, the charge across $C2$ and $C3$ is equal. Where $\Delta V$ is the "total potential" = $100V$. After the switch is flipped to the right, $C1$ discharges some of its potential to $C2$ and $C3$, determined by their capacitances.Īll three capacitors are connected in series, so $$C_)\cdot \Delta V$. Before the switch is closed, all the potential from the $100$ V battery is stored into $C1$. I want to make sure my understanding of this problem is clear. Initially the switch is connected to the left side, and the question is what are the charge and potential difference across each capacitor ( capacitance is in $\mu F$ ).
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