If you're ready, we'll do some examples of circuits that combine series and parallel elements in the same loop. They can take a few more steps – and those have to be done in the right order, but the logic is really just the same.
The approach here is to first solve for the total resistance of each parallel branch, adding resistors 1 & 2, and 3 & 4: to get 200 Ω and 300 Ω, respectively. From that we can reduce the complexity of the circuit (we always come back to the original version at the end):
Now we can find the total resistance of that parallel element:
So the total resistance is 120 Ω, and we have our final simplified equivalent circuit,
... from which we can calculate the current through the battery (I say "through the battery" because it will be different in the branches of the parallel paths once we work our way back to there).
Now we can step back and calculate the current in each arm of the parallel paths. I'll label the upper arm I200 and the lower I300, using the total series resistance of each as a label to keep things straight (you could also use "upper" & "lower" ... or whatever you want).
The total currents, calculated differently, agree. Ahhh .... science!
Finally, we calculate the voltage drops in the resistors of each arm using the appropriate current for each branch, 0.05A for the top, 0.033A for the bottom. They are
And that's the whole circuit – everything (well, almost) we'd want to know about it.
First we'll reduce the parallel resistors to a single resistance:
That gives us the equivalent series circuit:
which we can reduce, by adding 33Ω + 20Ω = 53Ω, to get the equivalent simple circuit (in the previous section, we learned that this is called Thevenin's equivalent),
Now the total current is
Stepping back to the first equivalent circuit (the one with two resistors in series), we can calculate the voltage drop across the resistors. I'll call them Vpar, the voltage drop across our 33 Ω parallel element, and Vser the drops across the 20 Ω resistor:
Notice that the sum of the voltage drops is equal to the battery voltage. Now that we have the drop across the parallel resistors (remember that parallel resistors share a common voltage drop), all that's left is to calculate the current through them:
There's a little rounding error there, but the sum of currents calculated this way agrees with the total current we calculated above. As a general rule, you should keep as many digits as possible (e.g. by not clearing your calculator) during the calculation and round at the end. Only worry about significance of digits when you report the result.
To calculate all currents and voltage drops for this circuit, we'll follow this road map:
First we'll convert each of the parallel elements to single equivalent resistances. I'll call them R1-2 and R3-4 after the labeled resistors.
Now we have this equivalent circuit:
... which we can use to calculate the overall resistance of the circuit,
and the Thevenin's equivalent circuit looks like this:
Then, as usual, we calculate the overall current flowing through the circuit using I = V/R:
Stepping back to the previous series equivalent circuit, we can use Ohm's law calculate the voltage drops across each of our parallel elements:
Now it's just a matter of using the appropriate voltage drops to calculate the currents through each resistor:
The total current here has a bit of roundoff error, but that's typical. The currents in resistors 3 & 4 are:
Solve for all voltage drops and currents everywhere in these DC circuits:
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