Question 496: Electrostats & Capacitance

Question

Three condensers of capacities ' $\mathrm{C}_1$ ', ' $\mathrm{C}_2$ ', ' $\mathrm{C}_3$ ' are connected in series with a source of e.m.f. ' $V$ '. The potentials across the three condensers are in the ratio

Accepted Answer

When capacitors are connected in series, they all carry the same charge. This is because the charge that is stored on one plate of a capacitor must equal the charge on the adjacent plate of the next capacitor. Therefore, if you consider the voltage across each individual capacitor, the relation can be described as follows:
For capacitors $ C_1 $, $ C_2 $, and $ C_3 $ with potentials $ V_1 $, $ V_2 $, and $ V_3 $ respectively, we have:
$  Q = C_1 V_1 = C_2 V_2 = C_3 V_3  $
Solving for the voltage, we get:
$  V_1 = \frac{Q}{C_1}, \quad V_2 = \frac{Q}{C_2}, \quad V_3 = \frac{Q}{C_3} $
Therefore, the potential (voltage) ratios will be inversely proportional to the capacitances:
$  V_1 : V_2 : V_3 = \frac{1}{C_1} : \frac{1}{C_2} : \frac{1}{C_3}  $
This relationship illustrates that the capacitor with the smallest capacitance will have the largest voltage across it, and vice versa. Hence, the potentials across the capacitors are in the ratio $\frac{1}{\mathrm{C}_1}: \frac{1}{\mathrm{C}_2}: \frac{1}{\mathrm{C}_3}$.

Answer

$1: 1: 1$

Answer

$\mathrm{C}_1: \mathrm{C}_2: \mathrm{C}_3$

Answer

$\mathrm{C}_1^2: \mathrm{C}_2^2: \mathrm{C}_3^2$

Answer

$\frac{1}{\mathrm{C}_1}: \frac{1}{\mathrm{C}_2}: \frac{1}{\mathrm{C}_3}$

Electrostats & Capacitance

Three condensers of capacities ' \mathrm{C}_1 ', ' \mathrm{C}_2 ', ' \mathrm{C}_3 ' are connected in series with a source of e.m.f. ' V '. The potentials across the three condensers are in the ratio

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