Irodov 3.114: The Smaller Capacitor Breaks First – Amit & Deepali Sinha Foundation

Problem Irodov 3.114

One capacitor, of capacitance $C_1 = 1.0$ $\mu$ F, can stand at most the voltage $V_1 = 6.0$ kV; a second, of capacitance $C_2 = 2.0$ $\mu$ F, can stand at most $V_2 = 4.0$ kV. The two are joined in series. What is the largest voltage the pair can withstand?

The two capacitors in series, each marked with its capacitance and the largest voltage it can stand.

Given

$C_1 = 1.0~\mu\text{F}$
$V_1 = 6.0$ kV
$C_2 = 2.0~\mu\text{F}$
$V_2 = 4.0$ kV
capacitors uncharged before connection (assumed)

Find

the largest voltage

V

across the series pair

Solution

Step 1In series both capacitors carry the same charge

The inner plates of the two capacitors and the wire between them form one isolated conductor. Starting uncharged, it stays neutral, so the charge $q$ that the outer terminals put on the outer plates is matched on both capacitors. Each voltage is then the charge over its own capacitance:

U_1 = \frac{q}{C_1}, \qquad U_2 = \frac{q}{C_2} \tag{1}

Step 2The charge each capacitor can bear

A capacitor fails when its own voltage exceeds its rating. Through (1) each rating becomes a ceiling on the common charge:

\begin{aligned} q &\le C_1V_1 = (1.0~\mu\text{F})(6.0~\text{kV}) = 6.0~\text{mC}\\ q &\le C_2V_2 = (2.0~\mu\text{F})(4.0~\text{kV}) = 8.0~\text{mC} \end{aligned}

Both conditions hold only below the smaller ceiling, which here belongs to the first capacitor:

q_{\mathrm{max}} = C_1V_1 = 6.0~\text{mC}

(2)

Step 3The voltage of the pair at that charge

The total voltage is the sum of the two from (1), evaluated at (2):

V = \frac{q_{\mathrm{max}}}{C_1} + \frac{q_{\mathrm{max}}}{C_2} = C_1V_1\left(\frac{1}{C_1} + \frac{1}{C_2}\right)

Multiplying out, and with the numbers $C_1/C_2 = 0.50$ and $V_1 = 6.0$ kV:

V = (6.0~\text{kV})(1 + 0.50)

V = V_1\left(1 + \frac{C_1}{C_2}\right) = 9.0~\text{kV}

(3)

Check

At $V = 9.0$ kV the charge is $6.0$ mC, so $C_1$ holds $6.0~\text{mC}/1.0~\mu\text{F} = 6.0$ kV, exactly its limit, and $C_2$ holds $6.0~\text{mC}/2.0~\mu\text{F} = 3.0$ kV, safely below its $4.0$ kV; the two add to $9.0$ kV. Had the first capacitor not been the weaker one, the formula would have exchanged roles: the general answer is the smaller of $V_1(1 + C_1/C_2)$ and $V_2(1 + C_2/C_1)$ , here $9.0$ kV against $12$ kV.

Remark

The pair withstands $9.0$ kV, less than the $10$ kV sum of the two ratings: series capacitors share the voltage by charge, not by strength, so one of them reaches its limit while the other still has margin. A string of capacitors on a high-voltage line is graded for this reason, each unit chosen (or shunted by a resistor) so that all reach their limits together.