Question

a circuit in which a capacitor of capacitance 4700microfarad is to be connected across a potential difference of 9v.He has 4700microfarad 6v capacitor available. draw the diagram to show how the 4 capacitor could be used for this purpose

Answer 1

Given physical quantities

$C_1$ = 4700 mkF

$\varphi_1$ = 9v

$C_2$ = 4700 mkF

$\varphi_2$ = 6v

$q_t = q_! + q_2 = C_!\varphi_1 + C_2\varphi_2 =$ 4700 mkF * 9v + 4700 mkF * 6v = 4700 mkF * 15v

$\varphi_t = \varphi_1 + \varphi_2 =$ 15v

first diagramm

The next diagram for this purpose

$C_1 = C_2 = C_3 = C_4 = \frac{C_t}{4} = \frac{q_t}{4U_t} = \frac{4700*15}{4*15}$ = 1175 mkF

$q_1 = q_2 = q_3 = q_4 = \frac{q_t}{4} = \frac{4700*15*10^{-6}}{4}$C = 17625 mkC