Question

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You have three capacitors, with capacitances C = 2.0 muF, C2 = 3.0 muF, and Cz = 6.0 muF. (a) What is the maximum capacitance you can achieve using these three capacitors? Why is that the maximum value? Submission type: Submit entries below as answer to receive credit Save entries below (not submitted for credit yet) Rich formatting » Submit Answer Tries 0/3 (b) What is the minimum capacitance you can achieve using these three capacitors? Why is that the minimum value? Submission type: Submit entries below as answer to receive credit Save entries below (not submitted for credit yet) Rich formatting »

Answer 1

Given C1 = 2$\mu$F, C2 = 3$\mu$F and C3 = 6$\mu$F.

(a) The maximum capacitance that can be achieved using these three capacitors is when these capacitors are connected in parallel. That maximum capacitance is,

mar С1 + C2 + C3 = 2+3+6 = 11pF

So the maximum capacitance that can be achieved using these capacitors is 11$\mu$F.

In parallel combination of capacitors, each capacitor is connected directly to the voltage source just as if it were all alone. So the total capacitance in parallel is just the sum of the individual capacitances. Henece maximum capacitance.

(b) The minimum capacitance that can be achieved using these three capacitors is when these capacitors are connected in series. That minimum capacitance is,

+ + man C1 C2 C3

$\frac{1}{C_{min}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{1}{2}+\frac{6+3}{6\times 3}=\frac{1}{2}+\frac{1}{2}=1\mu F$

Cmin 1μF

In series combination of capacitors, voltage dividion occurs. So the total capacitance in series is less than the individual capacitances. Henece minimum capacitance.

(c) The capacitors are connected in parallel and the the potential difference applied across the combination is V = 510V.

Since the capacitors are connected in parallel, the voltage across C1 is,

$V_{1}=V=510V$

The charge on C1 is,

$Q_{1}=C_{1}V_{1}=2\times 10^{-6}\times 510=1.02\times 10^{-3}C=1.02mC$

Since the capacitors are connected in parallel, the voltage across C2 is,

$V_{2}=V=510V$

The charge on C2 is,

$Q_{2}=C_{2}V_{2}=3\times 10^{-6}\times 510=1.53\times 10^{-3}C=1.53mC$

Since the capacitors are connected in parallel, the voltage across C3 is,

$V_{3}=V=510V$

The charge on C3 is,

$Q_{3}=C_{3}V_{3}=6\times 10^{-6}\times 510=3.06\times 10^{-3}C=3.06mC$