14.2. Circuit Basics http://www.ck12.org
par,weneedto f li pthe f ractiontogetR_total.
Now that we have three resistors in series (the two in parallel can be counted as one), we simply need to add them to
get the total resistance.
Rtotal=50Ω+^3375120 Ω+ 50 Ω= 128. 125 Ω
We can now solve for the current by using equation [7]
Itotal=
Vtotal
Rtotal=
120V
128. 125 Ω
=.94A
This is total net current through the circuit; it’s also the current across the 50Ωresistors, but not the ones connected
in parallel.
b) To find the power dissipated, we will use equation [4].
P=I×∆V=.94A×120V=112Wc) To find the current going through the 75Ωlight bulb, we must realize that a total of .94A goes through the two
light bulbs in parallel; according to the junction rule above, the currents across the two light bulbs must add to this.
Now we must find what proportion of the current the 75Ωlight bulb gets. To do this, we use our knowledge that
resistors in parallel have the same voltage drop and Ohm’s Law:
V 75 =I 75 × 75 Ω=V 45 =I 45 × 45 Ω
I 45 +I 75 =.94A, soI 45 =.94A−I 75Therefore,
I 75 × 75 Ω= (.94A−I 75 )× 45 ΩSolving for the needed current, we find:
I 75 ≈.35Ad) The brightness is determined by the power dissipated. More power means a brighter lightbulb. According to
equation [4], the power dissipated by a resistor can be written asI^2 R. Since we know the resistance of and current
across every resistor, we can simply calculate this quantity for each one. The order is 50Ω’s, 45Ω, then 75Ω. The
50 Ωis brighter than the 45Ωbecause the 50Ωgets considerably more current.
e) When the bulbs are wired entirely in parallel, the voltage drops across them will be the same. SinceP=IV, the
way to determine the brightest bulb is to look at the currents across them, which will be inversely related with their
resistances. So, the bulb with the lowest resistance will be the brightest, the one with the second lowest resistance
will be second, and so on. Therefore the order is 45Ω, 50Ω, and finally 75Ω.