Parallel Circuit Resistance
Equivalent Resistance for Resistors in Parallel:
1 / R = 1 / R1 + 1 / R2 + 1 / R3 +...

Three resistors, 60 Ω, 30 Ω, 20 Ω, are connected
in parallel across a 90V battery.
a)
Draw a schematic
b)
Find the current through each branch of the
circuit
c)
Find the equivalent resistance of the circuit
d)
Find the current through the battery
Answers
a)
b) 1.5A, 3A, 4.5A
c) Reff = 10Ω
d) I = 90V/10Ω = 9A
More Practice

a)
b)
c)
A circuit contains six 240-Ω lamps (60-W
bulbs) and a 10.0-Ω heater connected in
parallel. The voltage across the circuit is
120V. What is the current in the circuit…
when four lamps are turned on?
when all lamps are on?
when six lamps and the heater are
operating?
Answers
a) 2A
b) 3A
c) 15A **
**Note that the current changes so drastically when the
10.0 Ω heater is on because it’s the path of least
resistance (how poetic)…more coulombs can move
through the circuit every second (C/s = A) because there
is a path to take with little resistance working against
them. Woot! Woot!
Series vs. Parallel

Think of equivalent resistance like cars on a highway. On
a single lane highway with few exits (series), a car crash
(resistance) up ahead affects every car on the highway.
All cars have to pass that point so rate of movement
(cars/second or C/s) slows down. However, if there are
several lanes or exits you can follow to get out of the
traffic and get to the same place, wouldn’t you use
them? Well, so would charged particles! The car crash
will effect your rate less if you have another path you
can take. Especially if the other cars are also taking
alternate routes. Fewer cars per area = less traffic… or
in our case less effective resistance!
Parallel
Series