Question

A finite ladder is constructed by connecting several sections of 2 µF, 4 µF capacitor combinations as shown in figure (31-E18). It is terminated by a capacitor of capacitance C. What value should be chosen for C, such that the equivalent capacitance of the ladder between the points A and B becomes independent of the number of sections in between ?
Figure

Answer 1

The Value of equivalent capacitance of the ladder between the points A and B becomes independent of the number of sections in between is equal to 4 µF.

Explanation:

Considering a finite circuit as shown in below

From the circuit Capacitance C and 4 µF are in series  

We know that when capacitors are connected in series the total equivalent capacitance is the ratio of sum of capacitors to the product of the capacitors.

$c=c_{1}+2 \mu f \Rightarrow \frac{4 \times C}{4+C}+2 \Rightarrow \frac{4 C+8+2 C}{4+C}=C \Rightarrow 4 C+8+2 C=4 C+C^{2}=C^{2}-2 C-8=0 C=\frac{2 \pm \sqrt{4+4 \times 1 \times 8}}{2}=\frac{2 \pm \sqrt{36}}{2}=\frac{2 \pm 6}{2} C=\frac{2+6}{2}=4 \mu \mathrm{f} \therefore The value of C is 4 uf$

Hence the value of equivalent capacitance of the ladder between the points A and B becomes independent of the number of sections in between is equal to 4 µF.

Answer 2

Answer:

Here. In connection 4uf and C are in series.

• C1=4×C/4+c

Then, c1 and 2uf are in parallel connection.

C=c1+2uf

=4×c/4+c + 2

C=4c+8+2c/4+c

4c+c^2=4c+8+2c

0=c^2-2c-8

C=2+or-√4×4×1×8/2

C=2+or-√36/2

C=2+or-6/2

Ans. C=4uf

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