Question

In the given network of capacitors as shown in the figure, given that $C_1=C_2=C_3$$=400\text{pF}$ and $C_4=C_5=C_6=200\text{pF}$. The effective capacitance of the circuit between $\text{X}$ and $\text{Y}$ is.

• A. $810\text{pF}$
• B. $205\text{pF}$
• C. $600\text{pF}$
• D. $410\text{pF}$

Answer 1

The correct option is D $410\text{pF}$

Let us use the technique of succesive reduction.



Here, in thefigure $C^{\prime }=\frac{(C_3+C_4)\left(C_2\right)}{C_3+C_4+C_2}$

$\Rightarrow C^{\prime }=\frac{(400+200)\times \left(400\right)}{(400+200+400)}=\frac{\left(600\right)\left(400\right)}{1000}$

$\Rightarrow C^{\prime }=240\text{pF}$

Now, applying the formula and redrawing the circuit,


$C^{\prime \prime }=\frac{\left(400\right)\left(440\right)}{840}=209.5\text{pF}$

Now, $C^{\prime \prime }$ and $C_6$ are in parallel. So, the equivalent capacitance,

$C_{eq}=C^{\prime \prime }+C_6=209.5+200=409.5\text{pF}\approx 410\text{pF}$

Hence, option (d) is correct answer.