Answer to Question #195378 in Electric Circuits for kruthik

Question

2.  You need to decrease the capacitance of a part of a circuit from 4600 pF to 2500 pF. For this process, you can add an additional capacitor to the circuit. How that additional capacitor must be connected and what capacitance it should have? Assume that no other existing circuit elements were removed.

Answer 1

Solution:- Gives

C_net =4600 pF

C'_net =2500pF

Capacitor capacitance 4600pFto2500pF decrease then added additional capacitor C in series combination

Now

$\frac{1}{C'_{net}}=\frac{1}{C_{net}}+\frac{1}{C}\rightarrow(1)$

equation (1) put value

$\frac{1}{2500}=\frac{1}{4600}+\frac{1}{C}$

$\frac{1}{C}=\frac{1}{2500}-\frac{1}{4600}$

$C=\frac{2500\times4600}{4600-2500}\times10^{-12}F$

$C=5476\times10^{-12}F$

$C=5476pF$

Additional capacitor add in series combination C=5476pF