Question

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

Answer 1

Answer:

Capacitors 5 and 1 are in series.

Their equivalent capacitance, Ceq=C

Capacitors 5 and 1 are in series.

Their equivalent capacitance,

Ceq = (C1×C5)/(C1+C5)

= (2×2)/(2+2)

= 1 μF

∴ Ceq = 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

(2×2)/(2+2) = 1 μF

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

(2×2)/(2+2) = 1 μF

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

(2×2)/(2+2) = 1 μF

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

Answer 2

The Equivalent capacitance between the two points in the given circuit is 1 µF

Explanation:  

Step 1:

From the above figure we can say that Capacitors 5 & 1 are connected in series.

We know that the Equivalent capacitance of the two capacitors when connected in series is given by

                   $\mathrm{c}=\mathrm{C} 1 \mathrm{C} 2 / \mathrm{Cl}+\mathrm{C} 2$

Step 2:

Here C1 = 2 & C2 = 2, by substituting the values

                   $\mathrm{c}=2 \times 2 / 2+2=1 \mu \mathrm{F}$

Step 3:

Now, by solving the above circuit, considering the next connected capacitor this capacitor system is parallel to capacitor 6.  

Thus, the equivalent capacitance become $C=1+1=2 \mu F$

Step 4:

Next, the above capacitor system is in series with capacitor 2. So, the equivalent capacitance becomes

                   $2 \times 22+2=1 \mu F 2 \times 22+2=1 \mu F$

Step 5:

Likewise, by solving all the other capacitances we can say that the overall equivalent capacitance between the two points in the given circuit is equal to the 1 µF.