Explanation:
Given two capacitors 7.0 µF and 5.0 µF connected in series, let s calculate the effective capacitance.
[tex]\frac{1}{C_T} = \frac{1}{7.0} +\frac{1}{5.0}\\ \frac{1}{C_T} = \frac{5+7}{35} \\C_T = \frac{35}{12}\\ C_T = 2.9166\mu F[/tex]
Given
V = 5.0kV = 5000Volts
To get the total charge in the circuit, we will use the formula:
[tex]Q = C_TV\\Q = 2.9166\times 10^{-6} \times 5000\\Q = 0.014583 C\\Q = 1.4583 \times 10^{-2}C[/tex]
Since the same charge flows in a series connected capacitors, the charge on them in mC is 14.583mC
For the voltage across 7.0 µF:
[tex]V = \frac{Q}{C}\\V = \frac{0.014583}{0.000007} \\V = 2,083.28V[/tex]
Voltage across the 7.0 µF is 2,083.28V
The voltage across the 5.0 µF:
V = 5000 - 2,083.28
V = 2,916.72Volts
Voltage across the 5.0 µF is 2,916.72V
Answer:
The charge on the first capacitor is 14.59 mC
The voltage across the first capacitor is 2084 V
The charge on the second capacitor is 14.59 mC
The voltage across the second capacitor is 2918 V
Explanation:
Given;
first capacitor, C₁ = 7.0 µF
second capacitor, C₂ = 5.0 µF
potential difference, V = 5000 V
The equivalent capacitance is given by;
[tex]\frac{1}{C_{eq}}= \frac{1}{C_1} +\frac{1}{C_2}\\\\\frac{1}{C_{eq}}= \frac{C_1 + C_2}{C_1C_2}\\\\C_{eq} = \frac{C_1C_2}{C_1 +C_2}\\\\ C_{eq} = \frac{(7*10^{-6})(5*10^{-6})}{(7*10^{-6} \ +\ 5*10^{-6})}\\\\C_{eq} = 2.917*10^{-6} \ F[/tex]
The charge on the first capacitor is given by;
[tex]Q_1 = C_{eq} V\\\\Q_1 = 2.917*10^{-6} *5000\\\\Q_1 = 0.01459 \ C\\\\Q_1 = 14.59 \ mC[/tex]
The voltage across the first capacitor is given by;
[tex]V_1 = \frac{Q_1}{C_1} \\\\V_1 = \frac{0.01459}{7*10^{-6}}\\\\ V_1 = 2084.286 \ V[/tex]
V₁ = 2084 V
The charge on the second capacitor is given by;
[tex]Q_2 = C_{eq} V\\\\Q_2 = 2.917*10^{-6} *5000\\\\Q_2 = 0.01459 \ C\\\\Q_2 = 14.59 \ mC[/tex]
The voltage across the second capacitor is given by;
[tex]V_2 = \frac{Q_2}{C_2}\\\\ V_2 = \frac{0.01459}{5*10^{-6}}\\\\V_2 = 2918 \ V[/tex]
