You are given three capacitors, with capacitances 0.0078 µF, 12000 pF, and 0.027

Question

You are given three capacitors, with capacitances 0.0078 µF, 12000 pF, and 0.027 µF. Choosing from these three capacitors, you pick two and combine them to make a new capacitor. Given the possible combinations, what is the maximum and the minimum voltage difference would they make when charged to 120 µC? How would you connect them in each case?

Minimum______V Parallel? yes

Maximum______V Series? yes

*Note* Minimum isn't 3448 V and maximum isn't 26402 V. Although they should be around that range just from looking at the practice problem.

Two people have gotten it wrong so far so here's the practice problem and the correct answers if you would like to check your methods.

You are given three capacitors, with capacitances 0.0080 µF, 12000 pF, and 0.027 µF. Choosing from these three capacitors, you pick two and combine them to make a new capacitor. Given the possible combinations, what is the maximum and the minimum voltage difference would they make when charged to 120 µC? How would you connect them in each case?

Minimum = 3080 V in parallel

Maximum = 25000 V in series

Greatly appreciate it!

Explanation / Answer

here

Capaciatnce C and VOlatge are related as Q = CV

V = Q/C

so V is max when C is minimun and vice versa

so for Cmax, Cparalel = C1+C2

Cmax , let us choose two highest C's

so

Cmax = 0.0078 *10^-6 + 0.0027 *10^-6

Cmax = 105 uF

so Vmin = Q/C = 120 *10^-6/(105 *10^-6)

Volts this is obtianed from series l combination of Capacitiros

------------------------

Vmin = 1.14 Volts (in whne C in parallel)

Cmin = 1/Cs = 1/0.012 e -6 + 1/ 0.0078e -6

Cs = 4.72 nF

so

Vmax = 120 e -6/4.72 e -9

Vmax = 25423 Volts this is obtianed from parallel combination of Capacitiros

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