Ammeters & Voltmeters
An ammeter measures the current (Amps) in a wire (or a device). To do this, it
has to be inserted in the wire so that all the current flows through it (in series with
the device). The ideal ammeter has itself a very small resistance (approaching
zero) so that the voltage drop is negligibly small.
A voltmeter measures the voltage (Volts) across some other device (eg, a resistor
or a battery). To do this, it has to be connected across (in parallel) to that device.
The ideal voltmeter has itself a very large resistance (approaching infinite) so that
the current flowing through it is negligibly small.
RV
A
R(i)
V
RC – Circuits
I(t)
+
When the switch is open, there
is no charge on C, so Vc=0. At
t=0, when the switch is first
closed, this condition still
holds, since there was no time
yet for any charge to flow. So
the voltage drop across R is V0
and I(0)=V0/R.
R
V0
VC
C
If we wait for a long time, then
charge will have flowed to C
until the potential across C
equals that across the battery,
Vc=V0. The capacitor is now
fully charged, and there is no
current.
1
RC – Circuits II
What happens between t=0 and t very large?
Charge will flow at a rate of i(t)=dq/dt from the battery through the resistor to the
capacitor. The voltage across the capacitor will be V=q/C. We can write the circuit
equation (going around the loop) as:
q
dq q
= 0 ⇒ V0 − R − = 0 ⇒
C
dt C
V0
dq
1
+
q=
dt RC
R
V0 − iR −
This is a differential equation, which has a solution in the general form:
q = A + Be
( −t RC )
A and B are constants which we determine from the initial conditions: At t=0, q=0,
from which we learn that 0=A+B, so B=-A, and at t=∞ Vc=q/C=V0 so we must have
q=V0C=A+0, thus A=V0C. The solution for a charging capacitor is thus:
( −t )
q = V0C  1 − e RC 


RC-Circuits III
A charging capacitor thus has a time dependent charge
on it that follows the equation:
q
The voltage across the capacitor is just V(t)=q/C,
which gives:
To get the time dependent current we differentiate
q(t) wrt. t, which gives:
( t ) = V0C 1 − e(

V ( t ) = V0 1 − e

i (t ) =
−t
RC
)
( − t RC ) 




V0 ( − t RC )
e
R
The term RC has units of seconds, and is called the time constant, τ. After one time
constant (t= τ) the capacitor will have charged to 1-e-1=63% maximum value and the
current will have dropped from I0=V0/R to 1/e=37% of that value.
2
RC-Circuits IV
A discharging capacitor has fairly similar equations. The solutions are:
I(t)
q ( t ) = V0Ce
V ( t ) = V0 e
R
VC
C
i (t ) =
( −t RC )
( −t RC )
−V0 ( − t RC )
e
R
3