consider the figure and value of capacitances given in the below if a 4 volt battery is connected between points A and B total charge stored on the capacitors would be here C1=2UF C2=3UF C3=4UF and C4=5UF . The equivalent capacitance between A and B?
Answers
Answer:
The equivalent capacitance of a series combination of capacitors is given
\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+...
C
1
=
C
1
1
+
C
2
1
+...
The equivalent capacitance of a parallel combination of capacitors is given by
C=C_1+C_2+...C=C
1
+C
2
+...
The circuit configuration of the problem is shown in the figure.
Here C_1C
1
and C_2C
2
are in series. The equivalent capacitance is
\frac{1}{C_{12}}=\frac{1}{C_1}+\frac{1}{C_2}\\ \frac{1}{C_{12}}=\frac{1}{1}+\frac{1}{2}\\ C_{12}=0.667\mu F
C
12
1
=
C
1
1
+
C
2
1
C
12
1
=
1
1
+
2
1
C
12
=0.667μF
Again C_3C
3
and C_4C
4
are in series. The equivalent capacitance is
\frac{1}{C_{34}}=\frac{1}{C_3}+\frac{1}{C_4}\\ \frac{1}{C_{34}}=\frac{1}{3}+\frac{1}{4}\\ C_{34}=1.71\mu F
C
34
1
=
C
3
1
+
C
4
1
C
34
1
=
3
1
+
4
1
C
34
=1.71μF
Now C_{12}\ and\ C_{34}C
12
and C
34
are in parallel. The equivalent capacitance is
C=C_{12}+C_{34}\\ C=0.667+1.71=2.38\mu FC=C
12
+C
34
C=0.667+1.71=2.38μF
Therefore the capacitance between points A and B is 2.38\mu F2.38μF
The total charge in the circuit is given by
Q=CV=2.38\mu F\times 14V=33.32\mu CQ=CV=2.38μF×14V=33.32μC
The charge on the upper branch i.e, in the capacitors C_1\ and \ C_2C
1
and C
2
is
Q_1=Q_2=C_{12}V=0.667\mu F\times 14V=9.34\mu CQ
1
=Q
2
=C
12
V=0.667μF×14V=9.34μC
[ Since in series combination, same current flows through the capacitors the charge stored in the capacitors is also the same. ]
The charge on the upper branch i.e, in the capacitors C_3\ and \ C_4C
3
and C
4
is
Q_3=Q_4=C_{34}V=1.71\mu F\times 14V=23.94\mu CQ
3
=Q
4
=C
34
V=1.71μF×14V=23.94μC
Potential across C_1C
1
is
V_1=\frac{Q_1}{C_1}=\frac{9.34\mu C}{1\mu F}=9.34VV
1
=
C
1
Q
1
=
1μF
9.34μC
=9.34V
The potential across C_2C
2
is
V_2=\frac{Q_2}{C_2}=\frac{9.34\mu C}{2\mu C}=4.67VV
2
=
C
2
Q
2
=
2μC
9.34μC
=4.67V
The potential across C_3C
3
is
V_3=\frac{Q_3}{C_3}=\frac{23.94\mu C}{3\mu F}=7.98VV
3
=
C
3
Q
3
=
3μF
23.94μC
=7.98V
The potential across C_4C
4
is
V_4=\frac{Q_4}{C_4}=\frac{23.94\mu C}{4\mu F}=5.99VV
4
=
C
4
Q
4
=
4μF
23.94μC
=5.99V
Explanation:
5.99v is the capacitance