A 0.50-μF𝜇F and a 1.4-μF𝜇F capacitor (C1𝐶1 and C2𝐶2, respectively) are connected in series to a 17-VV battery.Part ACalculate the potential difference across each capacitor.
Question
A 0.50-μF𝜇F and a 1.4-μF𝜇F capacitor (C1𝐶1 and C2𝐶2, respectively) are connected in series to a 17-VV battery.Part ACalculate the potential difference across each capacitor.
Solution
In a series circuit, the total capacitance (Ct) is given by the formula:
1/Ct = 1/C1 + 1/C2
Substituting the given values:
1/Ct = 1/0.50 μF + 1/1.4 μF 1/Ct = 2 μF^-1 + 0.714 μF^-1 1/Ct = 2.714 μF^-1
Therefore, Ct = 1 / 2.714 μF^-1 = 0.368 μF
The total charge (Q) in the circuit is given by Q = Ct * V, where V is the voltage of the battery.
Substituting the given values:
Q = 0.368 μF * 17 V = 6.256 μC
In a series circuit, the charge across each capacitor is the same. Therefore, the potential difference (V) across each capacitor is given by V = Q / C.
For C1 (0.50 μF):
V1 = Q / C1 = 6.256 μC / 0.50 μF = 12.512 V
For C2 (1.4 μF):
V2 = Q / C2 = 6.256 μC / 1.4 μF = 4.468 V
Therefore, the potential difference across the 0.50-μF capacitor is 12.512 V and the potential difference across the 1.4-μF capacitor is 4.468 V.
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