Energy Loss When a Charged Capacitor Is Connected to Another
Physics · Electrostatic Potential And Capacitance · NEET
When a charged capacitor is connected to another (uncharged) capacitor, charge is shared but total energy always drops. Charge Q is conserved, so both reach a common potential V = (C1V1 + C2V2)/(C1 + C2), but some energy is always lost as heat in the wire and as tiny radiation. Memory hook: charge is saved, energy is not — the wire "eats" the difference. For two identical capacitors (one charged, one empty), exactly 50% of the energy is lost.
A capacitor charged to V is connected to an identical empty one. Charge Q is conserved and shared, so both settle at V/2. The stored energy falls from (1/2)CV^2 to (1/4)CV^2 — exactly half is lost as heat in the wire.
Your doubts, answered
Why is energy lost at all — nothing seems to be doing negative work?
When the two capacitors are joined, charge rushes from high potential to low potential. This moving charge is a current, and any real wire has some resistance, so I squared R heat is produced. Even a 'perfect' wire radiates a little energy as electromagnetic waves. So the loss is unavoidable — the final energy is always less than the initial energy. Charge is conserved, but energy is not.
Where does the lost energy go if the wire has zero resistance?
This is the classic trap. Even with an ideal zero-resistance wire, the loss is the SAME. The energy leaves as electromagnetic radiation and as the oscillation dies out. Physics guarantees the answer does not depend on the wire's resistance: the loss formula has no R in it. So do not assume 'no resistance means no loss'.
How do I find the common potential after connecting?
Use charge conservation. Total charge before = total charge after. V_common = (C1 V1 + C2 V2)/(C1 + C2). If the second capacitor is uncharged (V2 = 0), this becomes V_common = C1 V1/(C1 + C2). Both capacitors end at this same voltage because they are now in parallel.
What is the exact energy-loss formula?
For two capacitors at initial voltages V1 and V2, the loss is: Delta U = (1/2) [C1 C2/(C1 + C2)] (V1 - V2)^2. This is always positive (it depends on the SQUARE of the voltage difference), which is why energy is always lost. If V1 = V2, no charge flows and no energy is lost.
For two identical capacitors, why exactly 50% loss?
Take C1 = C2 = C, one charged to V and the other empty (V2 = 0). Common potential = V/2. Initial energy = (1/2)C V^2. Final energy = (1/2)(2C)(V/2)^2 = (1/4)C V^2 = half the initial. So exactly 50% is lost, no matter the numbers. This exact case appears in NEET again and again.
⚠️ The NEET trap ✗ Students think a resistanceless wire conserves energy, so final energy = initial energy (option 'remains the same'). ✓ Energy is ALWAYS lost when V1 is not equal to V2. The loss formula Delta U = (1/2)[C1C2/(C1+C2)](V1-V2)^2 has no R in it — the answer is independent of wire resistance. For identical capacitors (one charged, one empty), energy DECREASES by a factor of 2 (50% lost). 🧠 'Ideal wire = no resistance = no energy loss' — the most common NEET mistake here.
Real NEET questions
NEET 2017
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system:
A · Increases by a factor of 4
B · Decreases by a factor of 2 ✓
C · Remains the same
D · Increases by a factor of 2
Solution: Step 1: Battery removed, so charge Q is fixed (conserved). Step 2: A second identical capacitor in parallel doubles capacitance to 2C. Step 3: Common voltage = Q/(2C) = V/2 (halved). Step 4: U_initial = (1/2)C V^2. Step 5: U_final = (1/2)(2C)(V/2)^2 = (1/4)C V^2 = U_initial/2. Energy decreases by a factor of 2. Answer: B.
NEET 2022
A capacitor of capacitance C = 900 pF is charged fully by a 100 V battery B. Then B is disconnected and connected to another uncharged capacitor of capacitance C = 900 pF. The electrostatic energy stored by the system is:
A · 4.5 x 10^-6 J
B · 3.25 x 10^-6 J
C · 2.25 x 10^-6 J ✓
D · 1.5 x 10^-6 J
Solution: Step 1: Initial energy U_i = (1/2)C V^2 = (1/2)(900 x 10^-12)(100)^2 = 4.5 x 10^-6 J. Step 2: Connecting an equal uncharged capacitor gives the 50% case (identical capacitors, one empty). Step 3: U_final = U_i/2 = 2.25 x 10^-6 J. Answer: C.
NEET 2026
Two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:
A · 0.5 x 10^-6 J ✓
B · 1.0 J
C · 1.0 x 10^-6 J
D · 0.5 J
Solution: Step 1: Use the loss formula Delta U = (1/2)[C1 C2/(C1 + C2)](V1 - V2)^2 with V2 = 0. Step 2: C1 C2/(C1 + C2) = (200 x 200)/(400) pF = 100 pF = 100 x 10^-12 F. Step 3: Delta U = (1/2)(100 x 10^-12)(100)^2 = (1/2)(100 x 10^-12)(10^4) = 0.5 x 10^-6 J. Answer: A.
Solved Electrostatic Potential And Capacitance NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Is charge conserved when two capacitors are connected?
Yes. The total charge on the connected plates before joining equals the total charge after. This is the key equation used to find the common potential. Energy, however, is not conserved — some is always lost as heat and radiation.
What fraction of energy is lost for two identical capacitors?
Exactly 50% (half) is lost when a capacitor charged to V is connected to an identical uncharged one. The final energy is one-half of the initial energy, regardless of the actual C or V values.
Does the energy loss depend on the resistance of the connecting wire?
No. The loss formula Delta U = (1/2)[C1C2/(C1+C2)](V1-V2)^2 contains no resistance. A larger resistance only makes the loss happen slower, but the total energy lost is the same.
When is there no energy loss?
Only when both capacitors are already at the same potential (V1 = V2). Then no charge flows, no current, no heat — Delta U = 0.
What is the common potential formula?
V_common = (C1 V1 + C2 V2)/(C1 + C2). If the second capacitor is uncharged, V_common = C1 V1/(C1 + C2).
