Question : Two insulated metallic spheres of capacitances 3 µF and 5 µF are charged to potential of 300 and 500 volt respectively. They are connected by a wire. Calculate the common potential, charge on each sphere and the loss of energy.
Doubt by Mohini
Solution :
C1 = 3µF = 3×10-6 F
Solution :
C1 = 3µF = 3×10-6 F
C2 = 5µF = 5×10-6 F
V1 = 300 V
V2 = 500 V
Common Potential,
VC = (C1V1+C2V2)/(C1+C2)
= [(3×10-6)300+(5×10-6)500]/(3×10-6)+(5×10-6)
V1 = 300 V
V2 = 500 V
Common Potential,
VC = (C1V1+C2V2)/(C1+C2)
= [(3×10-6)300+(5×10-6)500]/(3×10-6)+(5×10-6)
= (900×10-6 + 2500×10-6)/(8×10-6)
= (3400×10-6)/(8×10-6)
= 3400/8
= (3400×10-6)/(8×10-6)
= 3400/8
= 425 V
Final Charge on each Capacitor
Q1' = C1V = 3µF×425 = 1275 µC
Q2' = C2V = 5µF×425 = 2125 µC
Initial Energy stored in the combination
Q1' = C1V = 3µF×425 = 1275 µC
Q2' = C2V = 5µF×425 = 2125 µC
Initial Energy stored in the combination
Ui = ½C1V1² + ½C2V2²
Ui = ½(3×10-6)(300)² + ½(5×10-6)(500)²
= 0.135 J + 0.625 J
= 0.76 J
Final Energy stored in the combination
Uf = ½C1VC² + ½C2VC²
= ½(3×10-6)(425)² + ½(5×10-6)(425)²
=½(10-6)(425)²[3+5]
=½(8)(10-6)(180625)
= 0.722500 J
=½(8)(10-6)(180625)
= 0.722500 J
Energy Loss = Uf-Ui
= 0.722500 - 0.76
= -0.037 J
= 0.722500 - 0.76
= -0.037 J
-ve sign shows that energy is being lost.
Energy lost = 0.037 J
Energy lost = 0.037 J
