Question

Two capacitors 8microF and 2microF are charged to the same potential of 100V separately. The charged capacitors are connected between 2 points A and B such that oppositely charged plate are joined to same point. What is potential difference between A and B ? Find the total energy stored in the 2 capacitors together before and after.
( If you know the answer, please explain the procedure and the formula used).​

Answer 1

Answer :

First of all we need to find charge on each capacitor$.$

★ Charge on capacitor A :

⭆ Q₁ = C₁ × V₁

⭆ Q₁ = 8 × 100

⭆ Q₁ = 800 μC

★ Charge on capacitor B :

⭆ Q₂ = C₂ × V₂

⭆ Q₂ = 2 × 100

⭆ Q₂ = 200 μC

Now lets find energy stored in each capacitor.

★ Energy stored in capacitor A :

⭆ U₁ = 1/2 C₁V₁²

⭆ U₁ = 1/2 (8)(100)²

⭆ U₁ = 40000 μJ

★ Energy stored in capacitor B :

⭆ U₂ = 1/2 C₂V₂²

⭆ U₂ = 1/2 (2)(100)²

⭆ U₂ = 10000 μJ

∴ Total energy before reconnection :

• U = U₁ + U₂
• U = 40000 + 10000
• U = 50000 μJ = 50mJ

After charging of each capacitor, both are connected in parallel such that oppositely charged plates are joined.

⧪ Common potential across this parallel connection is given by

➠ V' = Net charge/Net capacitance

➠ V' = (Q₁ - Q₂)/(C₁ + C₂)

➠ V' = (800-200)/(8+2)

➠ V' = 600/10

➠ V' = 60 volts

⧪ Energy of parallel combination after reconnection :

➠ U' = 1/2 (C₁ + C₂) V'²

➠ U' = 1/2 (8+2)(60)²

➠ U' = 5(3600)

➠ U' = 18000 μJ

➠ U' = 18 mJ

Answer 2

First of all we need to find charge on each capacitor

★ Charge on capacitor A :

⭆ Q₁ = C₁ × V₁

⭆ Q₁ = 8 × 100

⭆ Q₁ = 800 μC

★ Charge on capacitor B :

⭆ Q₂ = C₂ × V₂

⭆ Q₂ = 2 × 100

⭆ Q₂ = 200 μC

Now lets find energy stored in each capacitor.

★ Energy stored in capacitor A :

⭆ U₁ = 1/2 C₁V₁²

⭆ U₁ = 1/2 (8)(100)²

⭆ U₁ = 40000 μJ

★ Energy stored in capacitor B :

⭆ U₂ = 1/2 C₂V₂²

⭆ U₂ = 1/2 (2)(100)²

⭆ U₂ = 10000 μJ

∴ Total energy before reconnection :

U = U₁ + U₂

U = 40000 + 10000

U = 50000 μJ = 50mJ

After charging of each capacitor, both are connected in parallel such that oppositely charged plates are joined.

⧪ Common potential across this parallel connection is given by

➠ V' = Net charge/Net capacitance

➠ V' = (Q₁ - Q₂)/(C₁ + C₂)

➠ V' = (800-200)/(8+2)

➠ V' = 600/10

➠ V' = 60 volts

⧪ Energy of parallel combination after reconnection :

➠ U' = 1/2 (C₁ + C₂) V'²

➠ U' = 1/2 (8+2)(60)²

➠ U' = 5(3600)

➠ U' = 18000 μJ

➠ U' = 18 mJ