# Q CV = ```Circuit worksheet—Capacitors
Physics 104
We can treat circuits with capacitors in similar way to the circuits with resistors. We
still find that voltages across parallel elements are the same. Now instead of having
currents in series elements be the same, charges are the same. We can use a similar
methodology where we reduce the sets of capacitors to an equivalent capacitance and use
the ideas of series charge being the same and parallel voltages being the same. We can
use the following set of equations.
C1 In this circuit, C1 =2.0&micro;F, C2=1.0&micro;F, C3=3.0&micro;F, and V=20V.
Q = CV
1
Cseries
=
1
1
1
+
+
C1 C2 C3
V
C2
C3
C|| = C1 + C2 + C3
Begin by reducing C2 and C3 to an equivalent capacitance and redraw the circuit.
C1 Resolve the parallel set: C23 = C2 + C3
C23 = 1.0 + 3.0 =4.0 μF
C23
V
Next reduce the remaining capacitors to one and redraw the circuit.
Remaining capacitors
are in series:
C123
1
1
1
=
+
C123 C1 C23
2)( 4 )
(
C1C23
=
=
= 1.33 μF
C1 + C23 (2) + ( 4 )
V
Now use the idea that series capacitors have the same charge and parallel capacitors
have the same voltage to find the charge, Q, and the voltage, V, on each of the capacitors.
Charges on C1 and C23 are equal
and equal to that on C123
V−
Q1 = Q23 = Q123
⎛1
Q
Q
Q1 Q23
1 ⎞
−
= V − 123 + 123 = V − Q123 ⎜⎜ +
⎟⎟ = 0
C1 C23
C1
C23
⎝ C1 C23 ⎠
Q1 = Q23 = VC123 = (20V )(1.33μF ) = 26.6 μC
C123
Circuit worksheet—Capacitors
Physics 104
Q123
Q123
V1 =
&amp; V23 =
C1
C23
Voltages across C1 and C23
are not equal and sum to V
Q123 26.6 μC
=
= 13.3 V
C1
2.0 μF
Q
26.6 μC
= 6.65 V
V23 = 123 =
C23
4.0 μF
V1 =
Voltages across C2 and C3 are
equal and equal to that on C23
Charges on C1 and C23 are
not equal and sum to Q23
V1 + V23 = V
V23 = V2 = V3 = 6.65 V
Q123 =
Q2 = V23C2 = ( 6.65 V )(1 μF ) = 6.65 μC
Q3 = V23C3 = ( 6.65 V )(3 μF ) = 19.95 μC
V23
= Q2 + Q3
C23
Q2 + Q3 = Q23
Summarize the capacitance, charge and voltage of each capacitor C1 = 2.0 μF
Q1 = 26.6 μC
V1 = 13.3 V C1
V C2 = 1.0 μF
Q2 = 6.65 μC
V2 = 6.65 V
C2 C3 Since C2 = 1.0 = Q/V,
Q2 = V2 x 10-6
C3 = 3.0 μF
Q3 = 19.95 μC
V3 = 6.65 V
```