Question

In the figure shown, each capacitance $C_1$ is $6.0\mu F$ and each capacitance $C_2$ is $4.0\mu F$. The charge on $C_1$ nearest to $a$ when $V_{ab}=420V$ is :

• A. $840\mu C$
• B. $560\mu C$
• C. $600\mu C$
• D. $320\mu C$

Answer 1

The correct option is A $840\mu C$

Reduction of the farthest right leg yields:
$C={(\frac{1}{6.0\mu F}+\frac{1}{6.0\mu F}+\frac{1}{6.0\mu F})}^{-1}=2.0\mu F=\frac{C_1}{3}$

It combines in parallel with $C_2$ i.e.,
$C=4.0\mu F+2.0\mu F=6.0\mu F=C_1$

So, the next reduction is the same as the first : $C=2.0\mu F=\frac{C_1}{3}$
And the next is the same as the second, leaving $3C_1$'s in series. So, $C_{eq}=2.0\mu F=C_1/3$.

For the three capacitors nearest to points $a$ and $b$:
$Q_{C_1}=C_{eq}V=(2.0\times {10}^{-6}F)\left(420V\right)=8.4\times {10}^{-4}C=840\mu C$