Question

For the arrangement of capacitors as shown in the circuit, the effective capacitance between the points A and B is? (capacitance of each capacitor is 4 µF)
      

a.) \[4\mu F\]
b.) \[2\mu F\]
c.) $1\mu F$
d.) \[8\mu F\]

Answer 1

Hint: Whenever a complicated circuit is given as we have been asked to find out the result capacitance or resistance of the first step, we need to simplify the circuit in such a way that we can solve it in an easy manner. If possible, check if the circuit is making a bridge or not and if no bridge is found then try to solve the given resistance separately or in pairs.

Complete answer:
The above circuit can be redrawn as

The above circuit is a bridge circuit and No charge will flow in Capacitor 5
We can easily see that if we exclude the capacitor 5 then Capacitor 1 and capacitor 2 are in series with each other and capacitor 3 and capacitor 4 are in series with each other
And the resultant of capacitor 1 and 2 will be in parallel with the resultant of capacitor 3 and 4
Now we know that the formula of capacitor in series can be given as
$\dfrac{1}{{{C}_{12}}}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{C{}_{2}}$
On putting the values, we get
${{C}_{12}}=2\mu F$
Now the resultant of capacitor 3 and 4 will be given as
$\dfrac{1}{{{C}_{34}}}=\dfrac{1}{{{C}_{3}}}+\dfrac{1}{{{C}_{4}}}$
${{C}_{34}}=2\mu F$
 Now the resulting capacitance ${{C}_{12}}$and ${{C}_{34}}$are in parallel connection with each other
So, the resulting capacitance can be given as
${{C}_{eq}}=2\mu F+2\mu F$
${{C}_{eq}}=4\mu F$

Hence, we can say that option (a) is the correct answer to this question.

Note:
The formula for capacitance in series is the same as that of resistance in series and the formula for capacitance in parallel is the same as that of the formula for resistance in series. Both the capacitor and resistor have their formula opposite in series as well as in parallel connection.