Two capacitors when connected in series, the equivalent capacity is 4μF. when they are connected in parallel, equivalent capacity is 18μF. calculate capacities of capacitors.
Answers
Answer:
Explanation:
Three capacitors are connected in series. The magnitude of the charge on each plate is Q. (b) The network of capacitors in (a) is equivalent to one capacitor that has a smaller capacitance than any of the individual capacitances in (a), and the charge on its plates is Q.
We can find an expression for the total (equivalent) capacitance by considering the voltages across the individual capacitors. The potentials across capacitors 1, 2, and 3 are, respectively, 1=/1
V
1
=
Q
/
C
1
, 2=/2
V
2
=
Q
/
C
2
, and 3=/3
V
3
=
Q
/
C
3
. These potentials must sum up to the voltage of the battery, giving the following potential balance:
=1+2+3.
V
=
V
1
+
V
2
+
V
3
.
Potential V is measured across an equivalent capacitor that holds charge Q and has an equivalent capacitance S
C
S
. Entering the expressions for 1
V
1
, 2
V
2
, and 3
V
3
, we get
S=1+2+3.
Q
C
S
=
Q
C
1
+
Q
C
2
+
Q
C
3
.
Canceling the charge Q, we obtain an expression containing the equivalent capacitance, S
C
S
, of three capacitors connected in series:
1S=11+12+13.
1
C
S
=
1
C
1
+
1
C
2
+
1
C
3
.
Please refer the attachment!
