Introductory Physics Volume Two

3.5 Kirchhoff’s Rules 61
⊲ Problem 3.4
A 60 Watt light bulb is plugged into the wall receptacle which supplies
an electric potential of 120 Volts. How much current runs through the
bulb when you turn it on?
⊲ Problem 3.5
(a) Show that a resistor with a current I running through it has a power
of P = I 2 R.
(b) Show that a resistor with a voltage ∆V across it has a power of
P = (∆V ) 2 /R.
§ 3.5 Kirchhoff’s Rules
There are two theorems that are very useful in analysing a circuit.
The first theorem stems from the conservation of charge, that is, that
charge is neither created nor destroyed in a circuit. Consider a junction
where a number of elements come together.
Since the current does not build up at the junction the sum of current
going into the junction must be equal to the sum of the current going
out of the junction. In the case pictured above: I 1 + I 4 = I 2 + I 3 .
Theorem: Kirchhoff’s Junction Rule
The sum of the currents into a junction is equal to the sum of the
current out of a junction. ∑
Iin = ∑ I out
The second theorem stems from the conservation
of energy. Since the electric potential is the potential
energy per charge that is due to the electric field in
the system, if a charge moves around a loop in a circuit
and comes back to where it started it must be at
the same electric potential as when it started. Thus
if we add the electric potential differences of all the
elements that are crossed as you go around any loop
in the circuit the sum must be zero.