Question

In the figure shown, each capacitance $C_1$ is $6.0\mu F$ and each capacitance $C_2$ is $4.0\mu F$. With $420V$ across $a$ and $b$, the value of $V_c-V_d$ is :

• A. $24.6V$
• B. $46.7V$
• C. $18V$
• D. $72V$

Answer 1

The correct option is B $46.7V$

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, $C_{eq}=2.0\mu F=\frac{C_1}{3}$.

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

$V_{cd}=\frac{1}{3}\left(\frac{420}{3}V\right)=46.7V$