Question

Calculate the equivalent capacitance between points A and B in the circuit below. If a battery of $10\,V$ is connected across A and B, calculate the charge drawn from the battery by the circuit.

Answer 1

Hint: here, in the given diagram two series combinations are connected parallel to each other. Therefore, we will use the formula of the capacitance to calculate the capacitance in both combinations. Then, we will use the formula of net capacitance to calculate the equivalent capacitance between points A and B.

Complete step by step answer:
The circuit given in the question represents the wheatstone bridge. A Wheatstone bridge is defined as an electrical circuit that is used to measure an unknown resistance which is done by balancing two legs of a bridge circuit and one leg of which includes the unknown component. Here, ${C_1} = 10\,\mu F$ , ${C_2} = 20\,\mu F$ , ${C_3} = 5\mu F$ , ${C_4} = 10\mu F$ and ${C_5} = 50\,\mu F$ .
Therefore, we can say that in the capacitance ${C_5} = 50\,\mu F$ , there will be no passage of current.
Now, we can see in the diagram that there are two series combinations that are connected parallel to each other.
Therefore, the net capacitance of the upper series combination consisting of capacitors ${C_1}$ and ${C_2}$ is shown below
$\dfrac{1}{{{C_{1,2}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$
Now, putting the values of ${C_1}$ and ${C_2}$ , we get
$\dfrac{1}{{{C_{1,2}}}} = \dfrac{1}{{10}} + \dfrac{1}{{20}}$
$ \Rightarrow \,\dfrac{1}{{{C_{1,2}}}} = \dfrac{{2 + 1}}{{20}}$
$ \Rightarrow \,\dfrac{1}{{{C_{1,2}}}} = \dfrac{3}{{20}}$
$ \Rightarrow \,{C_{1,2}} = \dfrac{{20}}{3}$
Also, the net capacitance of the lower series combination consisting of capacitors ${C_3}$ and ${C_4}$ is given below
$\dfrac{1}{{{C_{3,4}}}} = \dfrac{1}{{{C_3}}} + \dfrac{1}{{{C_4}}}$
Now, putting the values of ${C_3}$ and ${C_4}$ , we get
$\dfrac{1}{{{C_{3,4}}}} = \dfrac{1}{5} + \dfrac{1}{{10}}$
$ \Rightarrow \,\dfrac{1}{{{C_{3,4}}}} = \dfrac{{2 + 1}}{{10}}$
$ \Rightarrow \,\dfrac{1}{{{C_{3,4}}}} = \dfrac{3}{{10}}$
$ \Rightarrow \,{C_{3,4}} = \dfrac{{10}}{3}$
Now, the equivalent capacitance can be calculated by taking the sum of the net capacitance of both the series upper and lower series combination and is shown below
${C_{net}} = {C_{1,2}} + {C_{3,4}}$
$ \Rightarrow \,{C_{net}} = \dfrac{{20}}{3} + \dfrac{{10}}{3}$
$ \Rightarrow \,{C_{net}} = \dfrac{{30}}{3}$
$ \Rightarrow \,{C_{net}} = 10\,\mu F$
This is the required value.
Therefore, the equivalent capacitance between points A and B is $10\,\mu F$ .

Note:
Here we have not used the value of the fifth capacitor ${C_5}$ in the solution.
This is because in the case of Wheatstone bridge one component in the circuit is unknown, that is there will be no passage of current.
Therefore, in this case, there will be no passage of current through the capacitor ${C_5}$ .