Question

Find the capacitance of the combination shown in figure (31-E16) between A and B.

Answer 1

Capacitors 5 and 1 are in series.
Their equivalent capacitance, $C_{\mathrm{eq}}=\frac{C_1{C}_5}{C_1+C_5}=\frac{2\times 2}{2+2}=1\mathrm{\mu F}$
∴ C_eq = 1
Now, this capacitor system is parallel to capacitor 6. Thus, the equivalent capacitance becomes
1 + 1 = 2 μF

The above capacitor system is in series with capacitor 2. Thus, the equivalent capacitance becomes
$\frac{2\times 2}{2+2}=1\mathrm{\mu F}$

The above capacitor system is in parallel with capacitor 7. Thus, the equivalent capacitance becomes
1 + 1 = 2 μF

The above capacitor system is in series with capacitor 3. Thus, the equivalent capacitance becomes
$\frac{2\times 2}{2+2}=1\mathrm{\mu F}$

The above capacitor system is in parallel with capacitor 8. Thus, the equivalent capacitance becomes
1 + 1 = 2 μF

The above capacitor system is in series with capacitor 4. Thus, the equivalent capacitance becomes
$\frac{2\times 2}{2+2}=1\mathrm{\mu F}$

Hence, the equivalent capacitance between points A and B of the given capacitor system is 1 μF.