In the image to the left the potential (V) is analogous to height.  According to the Loop Rule the potential difference caused by the battery () must be compensated for by the potential drops across the two resistors (r and R).  Notice that the potential (V) starts at V_a and then returns again to V_a after resistor R.

Q:    "A battery has an initial internal resistance of 0.75 and an emf of 9 V.  It is placed a cross a 5 resistor and a 10 µF capacitor hooked up in parallel.
a)    After a the capacitor has charged, what is the current through the resistor?
b)    What is the charge on the capacitor?
c)    If the battery is disconnected, how long will it take the capacitor to reach one-third of its initial voltage?"

A:    First we will start off by drawing a picture of the circuit.

a)  After a the capacitor has charged, what is the current through the resistor?
After the capacitor has charged, there is no current passing through it.  This means the resistors (the internal resistance of the battery, r, and the resistor, R ) are connected in series.  Thus, the equivalent resistance of the circuit, R_eq , is the sum of the internal resistance of the battery, r, and the resistance of the resistor, R:

R_eq = r + R

Because the resistors are in series, the current in this circuit, I, is constant and equal to the emf, , over the equivalent resistance, R_eq:

I =  / R_eq

Now solve for I:
I   = / R_eq
    = 9 V / (.75 + 5 )
    = 1.57 A

The current through the battery is 1.57 A.

b)  What is the charge on the capacitor?
After the capacitor, C, is charged, the voltage across it is the same as the voltage across the resistor, R, (= IR).  Therefore, the charge, Q, on the capacitor, C, is:

Q = CV_R = C (IR) = 10 µF (1.57 A) (5 ) = 78.3 µC.

The charge on the capacitor is 78.3 µC.

c)  If the battery is disconnected, how long will it take the capacitor to reach one-third of its initial voltage?
In our picture, we switch the battery off with switch, S.  With the switch opened, the battery and its internal resistance drop out of the circuit.  We can then draw a new circuit, with the capacitor as the source of power.

Since the capacitor is discharging, the charge on it varies with time according to the equation q(t) = Q e ^{(- t / RC)}.  To find the time, t, it takes the capacitor to reach one-third of its initial capacity, we substitute
q(t) = Q/3 into the equation and solve for t:

q(t) = Q e ^{(- t / RC)}

Q/3 = Q e ^{(- t / RC)}

1/3 = e ^{(- t / RC)}

ln (1/3) = ln (e ^{(- t / RC)})

- ln 3 = - t / RC

t = ( ln 3 ) (RC)

t = (ln 3) (5 ) (10 µF)

t = 54.9 µsec

After 54.9 µsec, the capacitor has be one-third discharged.

Q:    If it is the current flowing through your body that determines the amount of damage done from a shock, why do the signs warn of high voltage and not high current?

A:    According to Ohm's Law ( I = V/R ), the current is proportional to voltage for a constant resistance.  Therefore, with the resistance of the human body given, the larger the voltage the larger the current passing through the body.   Contact with a high voltage source is thus more dangerous than with a low voltage source.