The equivalent capacitance between points A and B in the network shown in fig. is
Answers
Explanation:
Given:
Between points A and B, there are three capacitors connected in parallel which are connected to the fourth capacitor in series.
Formula Used:
When capacitors are connected in parallel, the equivalent capacitance is given by:
C=C_1+C_2+C_3+....C=C
1
+C
2
+C
3
+....
When capacitors are connected in series, the equivalent capacitance is given by:
C=(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+....)^{-1}C=(
C
1
1
+
C
2
1
+
C
3
1
+....)
−1
Calculations:
Between points A and B, there are three capacitors connected in parallel:
\begin{lgathered}C'=C+C+C\\C'=3C\end{lgathered}
C
′
=C+C+C
C
′
=3C
C' is connected in series with C. Equivalent capacitance between point A and B would be:
\begin{lgathered}C_e=(\frac{1}{C'}+\frac{1}{C})^{-1}\\C_e=(\frac{1}{3C}+\frac{1}{C})^{-1}\\C_e=\frac{3C}{4}\end{lgathered}
C
e
=(
C
′
1
+
C
1
)
−1
C
e
=(
3C
1
+
C
1
)
−1
C
e
=
4
3C