(i) The total capacitance for the circuit is 5 μF.
(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.
(iii) The charge stored in 3μF capacitor is 6 x 10⁻⁶ C.
Total capacitance of the circuit
The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.
C1 and C2 are in series
[tex]\frac{1}{C_{12}} = \frac{1}{C_1 } + \frac{1}{C_2} \\\\\frac{1}{C_{12}} = \frac{1}{4 } + \frac{1}{4} \\\\\frac{1}{C_{12}} = \frac{1}{2} \\\\C_{12} = 2 \ \mu F[/tex]
C1 and C2 are parallel to C3
[tex]C_{123} = C_{12} + C_3\\\\C_{123} = 2\ \mu F + 2\ \mu F \\\\C_{123} = 4 \ \mu F[/tex]
C(123) is series to C5 and C6
[tex]\frac{1}{C_{t} } = \frac{1}{C_{123}} + \frac{1}{C_5} + \frac{1}{C_6} \\\\\frac{1}{C_{t} } = \frac{1}{4} + \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{t} } = \frac{12}{24} \\\\\frac{1}{C_{t} } = \frac{1}{2} \\\\C_t = 2 \ \mu F[/tex]
C7 and C8 are in series
[tex]\frac{1}{C_{78}} = \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{78}} = \frac{2}{6} \\\\\frac{1}{C_{78}} =\frac{1}{3} \\\\C_{78} = 3 \ \mu F[/tex]
Total capaciatnce of the circuit
Ct + C(78) = 2 μF + 3 μF = 5 μF
Total charge stored in the circuit
The total charge stored in the capacitor is calculated as follows;
Q = CV
Q = (5 x 10⁻⁶) x (20)
Q = 1 x 10⁻⁴ C
Charge stored in 3μF capacitor
Q = (3 x 10⁻⁶) x (20)
Q = 6 x 10⁻⁶ C
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