

Guide 215a. Solving Multiloop Circuit Problems The method of solving problems below applies to multiloop circuits with a single source of emf and several resistors.
Problem: Find the potential difference across and the current through each resistor of the circuit shown below. Given values are listed to the right of Circuit Diagram 1.
Begin (as we've already done) by assigning each circuit component a unique symbol and listing the given values. We've also marked particular points of the circuit. This is for easy referencing in the discussion below. Finding the equivalent resistance The next step is to find the equivalent resistance of the circuit. You can then use that result to find the total current. In order to find the equivalent resistance, we're going to redraw the circuit in a series of steps to simplify it. Refer to Circuit Diagram 2 for the first step. Part of the process is determining which resistors are in parallel and which are in series. In this case, R_{3} and R_{4} are in parallel, because points b and c are connected by a wire, assumed to have 0 resistance, and are therefore at the same potential. Likewise, points e and d are at the same potential, although it's a different potential than points b and c. We redraw the circuit as shown below with R_{34} representing the combined resistance of R_{3} and R_{4}. Note that in Circuit diagram 2, points b and c have been collapsed to the single point b_{2}, and points e and d have been collapsed to e_{2}.
Now we can see that R_{1}, R_{2}, and R_{34} are in series as they are part of a single path, and the same current must pass through all of them. We can then represent the equivalent circuit in terms of a single resistance as shown in Circuit Diagram 3.
Substituting in the expression for R_{34}, we have R_{eq} = R_{1} + R_{2} + [1/R_{3} + 1/R_{4}]^{1}. Substituting given values yields the following:
Note that we express the resistances as reduced fractions in order to eliminate roundoff error in calculations to come. (This is not always feasible, depending on how simple the fractions are. Sometimes, it's easier to express results as decimals to one digit more than the number of significant digits.) Applying the loop rule to find the total current In Circuit Diagram 3, shown again to the right for reference, the circuit is reduced to a single loop. Let's define I_{b} as the representing the current in the loop. Traversing the loop counterclockwise, we have for the loop rule the following.
Now we substitute V_{eq} = I_{b}R_{eq} and solve for I_{b} to obtain:
Substituting values yields I_{b} = (13.0 V)/(26/5 _{}Ω) = 5/2 A. Finding the remaining currents and potential voltages We can now work backwards to find the currents in and the potential differences across the individual resistors. Refer to Circuit Diagram 2 to the right. Since there's a single loop, the same current passes through each of the resistors, R_{1}, R_{2}, and R_{34}. (This is an application of the loop rule with a single loop.) Thus, we can say:
Knowing the currents, we can solve for the voltages using V = IR.
As a check, the loop rule must be satifsied:
The loop rule checks out. We can move on to deal with R_{3} and R_{4} individually. Refer to Circuit Diagram 4 below. Note that we've added arrows and labels indicating the currents in the resistors.
First we apply the junction rule. We could use either point e or point b as the junction. For point e, the junction rule gives I_{2} = I_{3} + I_{4}. The total current entering the junction is equal to the total current leaving the junction. If we used point b, we'd have I_{3} + I_{4} = I_{1}. Note that these two results are identical since I_{2} = I_{1}. We have two unknowns in I_{2} = I_{3} + I_{4}. I_{2} is known but I_{3} and I_{4} are not. Thus, we need more information in order to determine the two unknown currents. We will get that information from application of the loop rule and V = IR. Now using V = IR, we have the following:
Uisng the fact that V_{3} = V_{4}, we can say that I_{3}R_{3} = I_{4}R_{4}. Next we use the loop rule and solve for I_{4}. (We could have solved for I_{3} instead. The goal is to eliminate one of the unknown currents.) This gives I_{4} = I_{2}  I_{3}. Substitute this into I_{3}R_{3} = I_{4}R_{4} and solve for I_{3}.
We substitute the result for I_{3} back into the junction rule to obtain I_{4}.
Knowing the currents, we solve for the voltages V_{3} and V_{4}.
As a check, we see that V_{3} = V_{4} as expected. Another check is to revisit a previous result obtained much earlier, V_{34} = 3.0 V. For yet one more check, I_{3} + I_{4} = 2.5 A = I_{2}= I_{1}. Summary All of the numerical results are listed in the table below in decimal form to display significant figures.
Here's a complete analysis for a more complex circuit.


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