Skip to content
Exercises · 2.6

Q.Three capacitors each of capacitance 9 pF9\ \text{pF} are connected in series.

(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V120\ \text{V} supply?
Uttarakhand UbseTextbookSubjective· 2mImportance★★★★★
29% · 16/56 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

For identical capacitors in series, the total capacitance is the individual value divided by the number of capacitors — here 3 pF3\ \text{pF}. Since they are identical, the supply voltage divides equally across each, giving 40 V40\ \text{V} per capacitor.

When capacitors are connected in series, the charge on each capacitor is the same, but the voltage across the combination splits across them. This is the mirror image of resistors in parallel — and it’s a common source of confusion.

The key idea: for nn identical capacitors in series, the total capacitance is simply C/nC/n, and the voltage across each is V/nV/n. Let’s see why.


1. Total capacitance in series

For capacitors in series, the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances:

1Ctotal=1C1+1C2+1C3\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

Here each C=9 pFC = 9\ \text{pF}, so:

1Ctotal=19+19+19=39=13\frac{1}{C_{\text{total}}} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3}

Therefore:

Ctotal=3 pFC_{\text{total}} = 3\ \text{pF}

Tip

For nn identical capacitors each of capacitance CC in series, the shortcut is Ctotal=C/nC_{\text{total}} = C/n. Here 9/3=3 pF9/3 = 3\ \text{pF} — no need to write reciprocals each time.


2. Voltage across each capacitor

When the series combination is connected to a 120 V120\ \text{V} supply, the same charge QQ appears on each capacitor. The voltage across a capacitor is V=Q/CV = Q/C.

Since the capacitors are identical, the charge QQ is the same for all, and so the voltage across each is also the same. The total voltage is the sum of the individual voltages:

V1+V2+V3=120 VV_1 + V_2 + V_3 = 120\ \text{V}

If each voltage is VV, then 3V=120 V3V = 120\ \text{V}, so:

V=40 VV = 40\ \text{V} …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.