Question a8e29

May 10, 2016

$\text{charge of 2} \mu F \rightarrow 12 \cdot {10}^{-} 6 C$

$\text{charge of 4} \mu F \rightarrow 24 \cdot {10}^{-} 6 C$

$\text{charge of 6} \mu F \rightarrow 36 \cdot {10}^{-} 6 C$

$\text{charge of circuit} \rightarrow 36 \cdot {10}^{-} 6 C$

Explanation:

$\text{first; let's find the total capacitances in circuit. }$

$\text{for capacitors "4 mu F and 2 mu F," total="4+2=6 mu F"(in parallel)}$

"for capacitors "6 mu F " and " 6 mu F" " 1/("total")=1/(6)+1/6

$\text{total capacitance=} \frac{6}{2} = 3 \mu F$

C=3*10^-6F" (total capacitance for circuit")#

$C = \frac{Q}{V}$

$Q = C \cdot V$

$Q = 3 \cdot {10}^{-} 6 \cdot 12$

$Q = 36 \cdot {10}^{-} 6 C \text{ (total charge for circuit)}$

$x + 2 x = 36 \cdot {10}^{-} 6$

$3 x = 36 \cdot {10}^{-} 6$

$x = {12.10}^{-} 6 C$

$\text{charge of 2} \mu F \rightarrow 12 \cdot {10}^{-} 6 C$

$\text{charge of 4} \mu F \rightarrow 24 \cdot {10}^{-} 6 C$

$\text{charge of 6} \mu F \rightarrow 36 \cdot {10}^{-} 6 C$

$\text{charge of circuit} \rightarrow 36 \cdot {10}^{-} 6 C$