Question
Obtain the equivalent capacitance of the network in figure. For a 300 V supply, determine the charge and voltage across each capacitor.
Solution
Given,
Capacitance, C1 = 100 pF
Capacitance, C2 = 200 pF
Capacitance, C3 = 200 pF
Capacitance, C4 = 100 pF
Capacitance C2 and C3 are in series.
Capacitance, C1 = 100 pF
Capacitance, C2 = 200 pF
Capacitance, C3 = 200 pF
Capacitance, C4 = 100 pF
Capacitance C2 and C3 are in series.
The circuit can be redrawn as
Now we see, C23 and C1 are in parallel combination,
On redrawing the circuit,
As seen in the above fig.
C123 and C4 are in series.
= 66.67 pF
Equivalent capacitance is given by C1234 = 66.67 pF
The given circuit may be re-drawn as shown in following figure.
Now,
Potential across C123 and C4 is given by,
Total voltage is equal to 300 V.
So, voltage across C4 is 200 V.
Therefore, the voltage across the combination of C1, C2 and C3 is 100 V.
Since C1 and C23 are in parallel, therefore, the voltage across C1 as well as across the series combination of C2 and C3 is 100 V.
Again, C2 and C3 are equal therefore, 100 V would be shared equally between C2 and C3.
V2 = 50 V and V3 = 50 V
Therefore,
Voltage across capacitor 1, V1= 100 V
Voltage across capacitor 2, V2= 50 V
Voltage across capacitor 3, V3= 50 V
Voltage across capacitor 4, V4 = 200 V
Charge across each capacitor is Q=CV
Therefore,
Charge across capacitor 1, Q1 =C1V1= 100 x 10–12 x 100 = 10–8 C
Charge across capacitor 2, Q2 =C2V2= 200 x 10–12 x 50 = 10–8C
Charge across capacitor 3, Q3 =C3V3= 200 x 10–12 x 50 = 10–8 C
Charge across capacitor 4, Q4 =C4V4= 100 x 10–12 x 200 = 2 x 10–8 C
Now we see, C23 and C1 are in parallel combination,
On redrawing the circuit,
As seen in the above fig.
C123 and C4 are in series.
= 66.67 pF
Equivalent capacitance is given by C1234 = 66.67 pF
The given circuit may be re-drawn as shown in following figure.
Now,
Potential across C123 and C4 is given by,
Total voltage is equal to 300 V.
So, voltage across C4 is 200 V.
Therefore, the voltage across the combination of C1, C2 and C3 is 100 V.
Since C1 and C23 are in parallel, therefore, the voltage across C1 as well as across the series combination of C2 and C3 is 100 V.
Again, C2 and C3 are equal therefore, 100 V would be shared equally between C2 and C3.
V2 = 50 V and V3 = 50 V
Therefore,
Voltage across capacitor 1, V1= 100 V
Voltage across capacitor 2, V2= 50 V
Voltage across capacitor 3, V3= 50 V
Voltage across capacitor 4, V4 = 200 V
Charge across each capacitor is Q=CV
Therefore,
Charge across capacitor 1, Q1 =C1V1= 100 x 10–12 x 100 = 10–8 C
Charge across capacitor 2, Q2 =C2V2= 200 x 10–12 x 50 = 10–8C
Charge across capacitor 3, Q3 =C3V3= 200 x 10–12 x 50 = 10–8 C
Charge across capacitor 4, Q4 =C4V4= 100 x 10–12 x 200 = 2 x 10–8 C
