Problem
Two capacitors of capacitance 8μF and 10μF are connected in series across a 180V power supply. They are then disconnected from the supply and reconnected in parallel with each other. Calculate the new potential difference across the combination and the charge on each capacitor.
Data
- Capacitor C1​=8μF
- Capacitor C2​=10μF
- Voltage V=180V.
Prerequisite Concepts
- Equivalent Capacitance in Series:
Ceq​1​=C1​1​+C2​1​
- Charge on Capacitors in Series:
Q=Ceq​⋅V
- Equivalent Capacitance in Parallel:
Ctotal​=C1​+C2​
- Voltage Across Parallel Combination:
Vnew​=Ctotal​Qtotal​​
Answer
Step 1: Calculate Equivalent Capacitance in Series
Ceq​1​=C1​1​+C2​1​
Ceq​1​=81​+101​
Take LCM:
Ceq​1​=8010+8​=8018​
Ceq​=1880​=4.444μF
Step 2: Calculate Total Charge in Series
Q=Ceq​⋅V
Q=4.444⋅180=800μC
Step 3: Calculate Total Charge in Parallel
When reconnected in parallel, the total charge becomes:
Qtotal​=Q+Q=2Q
Qtotal​=2⋅800=1600μC
Step 4: Calculate Total Capacitance in Parallel
Ctotal​=C1​+C2​
Ctotal​=8+10=18μF
Step 5: Calculate New Voltage
Vnew​=Ctotal​Qtotal​​
Vnew​=181600​=88.89V
Step 6: Calculate Charge on Each Capacitor
For C1​:
Q1​=C1​⋅Vnew​
Q1​=8⋅88.89=711.12μC
For C2​:
Q2​=C2​⋅Vnew​
Q2​=10⋅88.89=888.9μC
Final Answer
- New Potential Difference: 88.89V
- Charge on C1​: 711.12μC
- Charge on C2​: 888.9μC.