CCommons
Search✍️ WriteSign in
βš›οΈ NEET Physics❓ Question

Charged capacitor connected to an uncharged one, charge is conserved but energy drops. Where does it go?

Pooja Iyer12th Sample

12 Sept 2026 Β· 1 min read

Okay drama alert 🎭 this chapter has betrayed me. Capacitors, combinations and energy.

Setup from a question I did: a 2Β ΞΌF2\ \mu\text{F} capacitor is charged to 100100 V and then connected (positive plate to positive plate) to an uncharged 3Β ΞΌF3\ \mu\text{F} capacitor.

I did charge conservation: total charge =2Γ—10βˆ’6Γ—100=2Γ—10βˆ’4= 2\times10^{-6}\times100 = 2\times10^{-4} C, shared across a combined 5Β ΞΌF5\ \mu\text{F}, so common potential is V=40V = 40 V. Nice.

Then energy. Before: 12Γ—2Γ—10βˆ’6Γ—1002=10\frac12 \times 2\times10^{-6}\times100^2 = 10 mJ. After: 12Γ—5Γ—10βˆ’6Γ—402=4\frac12 \times 5\times10^{-6}\times 40^2 = 4 mJ. So 6 mJ just vanished?! Energy is supposed to be conserved. The answer key said "energy lost as heat", but there are no resistors, just wires. I even drew it.

Also is there a quick formula for this loss? My teacher mentioned something with (V1βˆ’V2)2(V_1 - V_2)^2 but I didn't copy it in time. And can someone clear up the series/parallel thing, which quantity is same in which combination? I keep swapping them.

WhatsAppPost on X

7 answers & replies

Sign in to join the conversation.

  • [+]Meera Patel11thSampleΒ· 1mo ago

    Your numbers are right. The general result for two capacitors C1C_1 (at V1V_1) and C2C_2 (at V2V_2) joined with like plates together:

    Common potential: V=C1V1+C2V2C1+C2V = \dfrac{C_1V_1 + C_2V_2}{C_1 + C_2}

    Energy loss: Ξ”U=C1C22(C1+C2)(V1βˆ’V2)2\Delta U = \dfrac{C_1 C_2}{2(C_1 + C_2)}(V_1 - V_2)^2

    Check with yours: 2Γ—32Γ—5Γ—10βˆ’6Γ—1002=0.6Γ—10βˆ’6Γ—104=6Γ—10βˆ’3\dfrac{2\times3}{2\times5}\times10^{-6}\times100^2 = 0.6\times10^{-6}\times10^4 = 6\times10^{-3} J = 6 mJ. It matches exactly. So the loss is always positive unless V1=V2V_1 = V_2.

    • [+]Pooja Iyer12thSampleΒ· 1mo agoAuthor

      YES the formula!!! Written in gold. Also the tank analogy is going to live in my head now. Thank you, you absolute gems.

  • [+]Harsh Gupta12thSampleΒ· 1mo ago

    Where does it go? Into the wire as heat while the charge rushes across, and some radiates away as electromagnetic waves. Even with an ideal zero-resistance wire the energy is not "gone", it's just not stored in the capacitors anymore. NCERT mentions this kind of loss with the heat in the connecting wires. Charge flows through a real wire, a real wire has resistance, and IΒ²R does the rest. Think of two connected water tanks with different levels, water sloshes around and the energy is lost to turbulence.

  • [+]Kabir Joshi11thSampleΒ· 1mo ago

    Actually if you put a purely ideal wire with no resistance you'd get an oscillating current, an LC-ish thing, and the energy would radiate. In real life it's heat. But if the question says "assume no resistance" in an exam the answer remains "energy lost as heat and radiation" - the formula is the same.

  • [+]Ananya Reddy11thSampleΒ· 1mo ago

    Series vs parallel, my colour-coded version:

    • Parallel: same potential difference on each capacitor. Charges add. Ceq=C1+C2C_{eq} = C_1 + C_2.
    • Series: same charge on each capacitor. Potential differences add. 1Ceq=1C1+1C2\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}.

    Mnemonic: capacitors do the opposite of resistors. Resistors add in series, capacitors add in parallel. I put both in the same table side by side so I can see the flip.

    • [+]Vivek Mishra12thSampleΒ· 1mo ago

      I always write the series formula as "C_eq = C1 C2 / (C1 + C2) for two" but in the test I used the same one for three capacitors and got it wrong. For three or more you need the full reciprocal sum, not product over sum.

  • [+]Aarav Kulkarni11thSampleΒ· 1mo ago

    Can I add: in series, the smallest capacitor has the biggest voltage across it (since V = Q/C and Q is same). Useful for the question "which one breaks down first".

More from Pooja Iyer

Keep reading