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Problem

Two capacitors of capacitance 8 μF8 \, \mathrm{\mu F} and 10 μF10 \, \mathrm{\mu F} are connected in series across a 180 V180 \, \mathrm{V} 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=8 μFC_1 = 8 \, \mathrm{\mu F}
  • Capacitor C2=10 μFC_2 = 10 \, \mathrm{\mu F}
  • Voltage V=180 VV = 180 \, \mathrm{V}.

Prerequisite Concepts

  • Equivalent Capacitance in Series:
1Ceq=1C1+1C2 \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}
  • Charge on Capacitors in Series:
Q=Ceqâ‹…V Q = C_{\text{eq}} \cdot V
  • Equivalent Capacitance in Parallel:
Ctotal=C1+C2 C_{\text{total}} = C_1 + C_2
  • Voltage Across Parallel Combination:
Vnew=QtotalCtotal V_{\text{new}} = \frac{Q_{\text{total}}}{C_{\text{total}}}

Answer

Step 1: Calculate Equivalent Capacitance in Series

1Ceq=1C1+1C2\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} 1Ceq=18+110\frac{1}{C_{\text{eq}}} = \frac{1}{8} + \frac{1}{10}

Take LCM:

1Ceq=10+880=1880\frac{1}{C_{\text{eq}}} = \frac{10 + 8}{80} = \frac{18}{80} Ceq=8018=4.444 μFC_{\text{eq}} = \frac{80}{18} = 4.444 \, \mathrm{\mu F}

Step 2: Calculate Total Charge in Series

Q=Ceq⋅VQ = C_{\text{eq}} \cdot V Q=4.444⋅180=800 μCQ = 4.444 \cdot 180 = 800 \, \mathrm{\mu C}

Step 3: Calculate Total Charge in Parallel

When reconnected in parallel, the total charge becomes:

Qtotal=Q+Q=2QQ_{\text{total}} = Q + Q = 2Q Qtotal=2⋅800=1600 μCQ_{\text{total}} = 2 \cdot 800 = 1600 \, \mathrm{\mu C}

Step 4: Calculate Total Capacitance in Parallel

Ctotal=C1+C2C_{\text{total}} = C_1 + C_2 Ctotal=8+10=18 μFC_{\text{total}} = 8 + 10 = 18 \, \mathrm{\mu F}

Step 5: Calculate New Voltage

Vnew=QtotalCtotalV_{\text{new}} = \frac{Q_{\text{total}}}{C_{\text{total}}} Vnew=160018=88.89 VV_{\text{new}} = \frac{1600}{18} = 88.89 \, \mathrm{V}

Step 6: Calculate Charge on Each Capacitor

For C1C_1:

Q1=C1⋅VnewQ_1 = C_1 \cdot V_{\text{new}} Q1=8⋅88.89=711.12 μCQ_1 = 8 \cdot 88.89 = 711.12 \, \mathrm{\mu C}

For C2C_2:

Q2=C2⋅VnewQ_2 = C_2 \cdot V_{\text{new}} Q2=10⋅88.89=888.9 μCQ_2 = 10 \cdot 88.89 = 888.9 \, \mathrm{\mu C}

Final Answer

  • New Potential Difference: 88.89 V88.89 \, \mathrm{V}
  • Charge on C1C_1: 711.12 μC711.12 \, \mathrm{\mu C}
  • Charge on C2C_2: 888.9 μC888.9 \, \mathrm{\mu C}.