Two capacitors of capatance 3mewF and 5 mewF are connected in series . calculate the equivalent capacitors . if the battery of emf 10V is connected across them. calculate the charge on each capacitor and also potenial difference across each capacitor
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C₁=3*10⁻⁶F
C₂=5*10⁻⁶F
and V=10volts
effective capacitance C=C₁C₂/C₁₊C₂=15/8*10⁻⁶F=1.875*10⁻⁶F
q=CV=1.875*10⁻⁶*10=1.875*10⁻⁵C
Potential difference across each capacitor
V₁=q/C₁=1.875*10⁻⁵/3*10⁶=6.2V
V₂=q/C₂=1.875*10⁻⁵/5*10⁻⁶=3.75V
Charge on each capacitor
q₁=C₁V=3*10⁻⁶*10=3*10⁻⁵C
q₂=C₂v=5*10⁻⁶*10=5*10⁻⁵C
C₁=3*10⁻⁶F
C₂=5*10⁻⁶F
and V=10volts
effective capacitance C=C₁C₂/C₁₊C₂=15/8*10⁻⁶F=1.875*10⁻⁶F
q=CV=1.875*10⁻⁶*10=1.875*10⁻⁵C
Potential difference across each capacitor
V₁=q/C₁=1.875*10⁻⁵/3*10⁶=6.2V
V₂=q/C₂=1.875*10⁻⁵/5*10⁻⁶=3.75V
Charge on each capacitor
q₁=C₁V=3*10⁻⁶*10=3*10⁻⁵C
q₂=C₂v=5*10⁻⁶*10=5*10⁻⁵C
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