Answer to Question #154835 in Electric Circuits for Asil

Question #154835

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

1
Expert's answer
2021-01-11T11:35:30-0500

Given physical quantities

C1C_1 = 4700 mkF

φ1\varphi_1 = 9v

C2C_2 = 4700 mkF

φ2\varphi_2 = 6v

qt=q!+q2=C!φ1+C2φ2=q_t = q_! + q_2 = C_!\varphi_1 + C_2\varphi_2 = 4700 mkF * 9v + 4700 mkF * 6v = 4700 mkF * 15v

φt=φ1+φ2=\varphi_t = \varphi_1 + \varphi_2 = 15v

first diagramm



The next diagram for this purpose


C1=C2=C3=C4=Ct4=qt4Ut=470015415C_1 = C_2 = C_3 = C_4 = \frac{C_t}{4} = \frac{q_t}{4U_t} = \frac{4700*15}{4*15} = 1175 mkF

q1=q2=q3=q4=qt4=4700151064q_1 = q_2 = q_3 = q_4 = \frac{q_t}{4} = \frac{4700*15*10^{-6}}{4}C = 17625 mkC





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