Question

I need help on #18 please.

3.15 uF 4.4 F 7.88 UF 누 3.56 UF 10.8 V 는

Answer 1

We have $\small C_{1}=4.4\mu f$ ,$\small C_{2}=3.15\mu f\, ,\, C_{3}=3.56\mu f,\, C_{4}=7.88\mu f$

First we will simply the circuit and find equivalent capacitance of $\small C_{2}=3.15\mu f\, and\, C_{3}=3.56\mu f\,$

since they are parallel to each other $\small C^{'}_{eq}= C_{2} + C_{3}=3.15+3.56=6.71\mu f$

Now all three capacitors $\small C^{'}_{eq},C_{1}, C_{4}$ are in series,hence have same charge in all of them.

NOW we calculate equivalent capacitance of complete circuit.

$\small 1/C_{eq}=1/C^{'}_{eq}+1/C_{1}+1/ C_{4} = (1/6.71)+(1/4.4)+(1/7.88)$

hence $\small C_{eq}=1.98\mu f$

since all capacitor having same charge,using Q=CV=1.98x10.8=$\small 21.4\mu C$

using same formula for $\small C=7.88\mu f$ , V=Q/C= 21.4/7.88= 2.72V

hence V = 2.72V