Consider a circuit with 2 identical bulbs wired in parallel. If we add another bulb in parallel,

explain the changes to the measured values of the circuit properties. (Be explicit – for
example, you could say the value doubles, but don’t just say the quantity increases.)
a. Voltage.
b. Current.
c. Equivalent resistance.
d. Power of the circuit.
e. The brightness of each bulb.

2 answers

Assuming battery or power line resistance is low the voltage will not change
However the current now has a third path to follow so the current will increase. If each resistor is R original resistance was R/2
new resistance = Rnew
Rnew = (1/3) R
i original = V/(R/2) = 2 V/R
i new = 3/2 of current with just 2 resistors in parallel = 3V/R

original power = V (2 V/R) = 2 V^2/R
end power = V(3 V/R) = 3 V^2/R
Voltage is the same for each bulb of course, until you blow the fuse :)