Two capacitors of capacitances 6 µF and 12 µF are connected in series with 12 V battery. Find the equivalent capacitance of the combination. Find the charge and the potential difference across each capacitor. Ans. (4 µF, 48 µC, 8 V, 4 V)
Data Given:
Capacitance of capacitor 1 = C₁ =6 µF = 6 𝐱 10⁻⁶ F
Capacitance of capacitor 2 = C₂ =12 µF = 12 𝐱 10⁻⁶ F
Potential Difference (Voltage) = V = 12 V
To Find:
(1) Equivalent Capacitor = Cₑ = ?
(2) Charge on each Capacitor = Q = ?
(3) Potential Difference (Voltage) across capacitor 1 = V₁ = ?
(4) Potential Difference (Voltage) across capacitor 2 = V₂ = ?
Solution:
(1) Equivalent Capacitor = Cₑ = ?
We have the formula Equivalent capacitor connected in series as
`\frac{1}{Cₑ}` = `\frac{1}{C₁}` + `\frac{1}{C₂}`
Now by putting the values
`\frac{1}{Cₑ}` = `\frac{1}{6 x 10⁻⁶ F }` + `\frac{1}{12 x 10⁻⁶ F }`
`\frac{1}{Cₑ}` = `\frac{2 + 1}{12 x 10⁻⁶ }` F
`\frac{1}{Cₑ}` = `\frac{3}{12 x 10⁻⁶ }` F
or
Cₑ = `\frac{12 x 10⁻⁶}{3}` F
Cₑ = 4 𝐱 10⁻⁶ F = 4 µF ---------------Ans. 1
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(2) Charge on each Capacitor = Q = ?
We know that
Q = CV
Q = 4 𝐱 10⁻⁶ F x 12 V
Q = 48 x 10⁻⁶ C
Q = 48 µC -------------------------Ans. 2
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(3) Potential Difference (Voltage) across capacitor 1 = V₁ = ?
We have
Q = C₁V₁
or
V₁ = `\frac{Q}{C₁}`
by putting values
V₁ = `\frac{48 x 10⁻⁶ C}{6 x 10⁻⁶ F }`
V₁ = 8 V -----------------------------Ans. 3
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(4) Potential Difference (Voltage) across capacitor 2 = V₂ = ?
We have
Q = C₂V₂
or
V₂ = `\frac{Q}{C₂}`
by putting values
V₂ = `\frac{48 x 10⁻⁶ C}{12 x 10⁻⁶ F }`
V₂ = 4 V ----------------------Ans. 4
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