14.6. Capacitor Example http://www.ck12.org
14.6 Capacitor Example
Question: Consider the figure above when switch S is open. Note that the power supply is set to 24 V.
a) What is the voltage drop across the 20Ωresistor?
b) What current flows through the 60Ωresistor?
c) What is the voltage drop across the 20 microfarad capacitor?
d) What is the charge on the capacitor?
e} How much energy is stored in that capacitor?
Answer:
a) When the capacitor is charged — in the steady state — no current flows across it, and we basically have a circuit
with two resistors in series. Accordingly, the voltage drop across the 20Ωresistor will be in the same proportion to
the net voltage across the circuit as its resistance is to the net resistance (see circuits chapter):
20 Ω
20 Ω+ 60 Ω=. 25
This means that the voltage drop across the resistor is
. 25 ×24V=6V
b) Since there is only one path for the current to take, its value is the same everywhere on the circuit; all we have to
do is find the total current. This will then also be the amount of current that flows through the 60Ωresistor. We can
find it by applying Ohm’s Law for the circuit:
Rtotal= 60 Ω+ 20 Ω= 80 ΩSince we have the total resistance and the total voltage, we can solve for the total current using Ohm’s law.
V=RI⇒I=V
R
=
24V
80 Ω
=.3A
The current flowing through the resistor is therefore.3A.
c) We can find the voltage drop across the 20μFcapacitor by realizing that the voltage drop across any parallel
paths in a circuit have to be equal; otherwise the loop rule would be violated. Therefore, the voltage drop across the
capacitor is the same as the voltage drop across the 60Ωresistor. We can find this analogously to how we found the
voltage drop across the other resistor:
60 Ω
20 Ω+ 60 Ω