Solving Multiloop Circuit Problems

Guide 21-5b. Solving Multiloop Circuit Problems

Here's a more involved problem than the previous one. Like the previous problem, however, there's only one battery. There are, of course, more resistors. However, the overall plan of the solution is the same.

Collapse the circuit down to simpler equivalent circuits by identifying which resistors are in parallel and which are in series.
Determine the equivalent resistance of the circuit.
Determine the total current provided by the battery.
Use a combination of loop and junction rules as well as V = IR to determine the current in and voltage across each individual resistor.

Problem: Find the potential difference across and the current through each resistor of the circuit shown below. Given values are listed to the

V_b	12 V
R₁	1 Ω
R₂	5 Ω
R₃	2 Ω
R₄	3 Ω
R₅	4 Ω
R₆	2 Ω

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, redraw the circuit in a series of steps in order to simplify it. We recommend doing this by hand until you feel familiar enough with the process to carry it out in your head. In order to simplify the circuit, you have to recognize which resistors are in parallel and which are in series. Note that R₅ and R₆ are in series, because the same current must pass through both resistors following the single route edc. We redraw the circuit as shown below with R_5,6 representing the combined resistance of R₅ and R₆. Note also that we've redrawn R₄ to make it clear that it is in parallel with both R₃ and R_5,6. We know they're in parallel because i) one end of all three resistors is connected to the same point e as seen in Circuit diagram 1 and are therefore at the same potential, ii) the other end of all three resistors is connected through points b and c, which are at the same potential. Thus, all three resistors R₃, R₄, and R_5,6 have the same potential difference across them. Note that in Circuit diagram 2, points b, c₁, and c₂ are at the same potential, while points e₁, e₂, and e₃ are also at the same potential (but different from b, c₁, and c₂).

	By the formula for resistors in series. R_5,6 = R₅ +R₆.
Circuit diagram 2

Now we can combine R₃, R₄, and R_5,6 as shown below.

	By the formula for resistors in parallel,
Circuit diagram 3

We now see that R₁, R_3,4,5,6, and R₂ are in series, because the current has only one path: feba. They can be combined into a single resistance which is the equivalent resistance, R_eq, of the entire circuit. We calculate that value to the right of the circuit.


Circuit diagram 4

Finally, we note the other combined resistances, because they'll come in handy in finding currents and potential differences.

R_5,6= 6 Ω
R_3,4,5,6= 1 Ω

Finding currents and potential differences

Once R_eq is known, find the total current, I_b, provided by the battery. All the currents are shown in Circuit diagram 5. First we see that the currents I₁ and I₂ must be the same as I_b, because R₁ and R₂ are in series with the battery. We also see that currents I₅ and I₆ are the same, because R₅ and R₆ are in series. On the other hand, currents I₃, I₄, and I₅, are, in general, different, because they branch off from the junction e. However, by the junction rule, we know that I₂ = I₃ + I₄ + I₅.

Circuit diagram 5

With these relationships in mind, let's proceed to work backwards in finding currents and potential differences. Examining Circuit diagram 4, we know that:

I_b = V_b/R_eq = (12 V)/(7 Ω) = 12/7 A.

Now let's move on to Circuit diagram 3. We see that R₁, R₂, R_3,4,5,6 and the battery are in series and must all have the same current. Therefore, I_b = I₁ = I₂ = I_3,4,5,6 = 12/7 A. We can now calculate the potential differences across the resistors.

V₁ = I₁R₁ = (12/7 A)(1 Ω) = 12/7 V
V₂ = I₂R₂ = (12/7 A)(5 Ω) = 60/7 V
V_3,4,5,6 = I_3,4,5,6R_,3,4,5,6 = (12/7 A)(1 Ω) = 12/7 V

As a check on our work so far, the loop rule tells us that the potential rise V_b across the battery should equal the sum of the potential drops across the resistors.

V₁ + V₂ + V_3,4,5,6 = (12/7 + 60/7 + 12/7) V = 84/7 V = 12 V = V_b.

We have left to deal with resistors R₃ to R₆ individually. Let's back up to Circuit diagram 2. We see that R₃, R₄, and R_5,6 must have the same potential difference across them. We've already calculated this value above as V_3,4,5,6.

Thus, V₃ = V₄ = V_5,6 = 12/7 V. We can use this value to find the currents through the resistors.

I₃ = V₃/R₃ = (12/7 V)/(2 Ω) = 6/7 A
I₄ = V₄/R₄ = (12/7 V)/(3 Ω) = 4/7 A
I_5,6 = V_5,6/R_5,6 = (12/7 V)/(6 Ω) = 2/7 A

Before going on, we can use the junction rule to provide another check on our work. The three currents above all branch off at the same point. The current going in to that point is I₂ = 12/7 A. The sum of 6/7, 4/7, and 2/7 is 12/7, so the junction rule checks.

The last thing to do is find the potential differences across R₅ and R₆ individually. We already know the current in these resistors, so the potential differences are:

V₅ = I₅R₅ = (2/7 A)(4 Ω) = 8/7 V
V₆ = I₆R₆ = (2/7 A)(2 Ω) = 4/7 V

As a final check, the sum of V₅ and V₆ should be 12/7 V, since we had calculated that previously as V_5,6. Note that we left the results as reduced fractions, since that eliminated any roundoff error that would result from using decimals. Of course, you may do your calculations in decimal form instead. If you do so, avoid rounding until the last step by carrying calculations in your calculator.