MODEL Assume an ideal battery, with no internal resistance, and ideal connecting wires.

VISUALIZE Figure 28.28 shows the circuit diagram. We’ll keep redrawing the diagram as we analyze the circuit.

SOLVE First, we break the circuit down, step by step, into one with a single resistor. FIGURE 28.29a shows this done in three steps. The final battery-and-resistor circuit is our basic circuit, with current

FIGURE 28.29 The step-by-step circuit analysis.

Full Alternative Text

The potential difference across the $400\text{Ω}$ resistor is $\text{Δ}{V}_{400}=$ $\text{Δ}{V}_{\text{bat}}=\epsilon =12\text{V}$.

Second, we rebuild the circuit, step by step, finding the currents and potential differences at each step. FIGURE 28.29b repeats the steps of Figure 28.29a exactly, but in reverse order. The $400\text{Ω}$ resistor came from two $800\text{Ω}$ resistors in parallel. Because $\text{Δ}{V}_{400}{=}_{}12\text{V,}$ it must be true that each $\text{Δ}{V}_{800}=12\text{V}$. The current through each $800\text{Ω}$ is then $I=\text{Δ}V\text{/}R=15\text{mA}$. The checkpoint is to note that $15\text{mA}+15\text{mA}=30\text{mA}$.

The right $800\text{Ω}$ resistor was formed by $240\text{Ω}$ and $560\text{Ω}$ in series. Because $I_{800}=15\text{mA,}$ it must be true that $I_{240}=I_{560}=15\text{mA}$. The potential difference across each is $\text{Δ}V=IR$, so $\text{Δ}{V}_{240}=3.6\text{V}$ and $\text{Δ}{V}_{560}=8.4\text{V}$. Here the checkpoint is to note that $3.6\text{V}+8.4\text{V}=12\text{V}=\text{Δ}{V}_{800}$, so the potential differences add as they should.

Finally, the $240\text{Ω}$ resistor came from $600\text{Ω}$ and $400\text{Ω}$ in parallel, so they each have the same 3.6 V potential difference as their $240\text{Ω}$ equivalent. The currents are $I_{600}=6\text{mA}$ and $I_{400}=9\text{mA}$. Note that $6\text{mA}+9\text{mA}=15\text{mA}$, which is our third checkpoint. We now know all currents and potential differences.

REVIEW We checked our work at each step of the rebuilding process by verifying that currents summed properly at junctions and that potential differences summed properly along a series of resistances. This “check as you go” procedure is extremely important. It provides you, the problem solver, with a built-in error finder that will immediately inform you if a mistake has been made.

MODEL Assume ideal batteries and ideal connecting wires.

VISUALIZE Figure 28.30 shows the circuit diagram. None of the resistors are connected in series or in parallel, so this circuit cannot be reduced to a simpler circuit.

SOLVE Kirchhoff’s loop law applies to any loop. To analyze a multi-loop problem, we need to write a loop-law equation for each loop. FIGURE 28.31 redraws the circuit and defines clockwise currents $I_1$ in the left loop and $I_2$ in the right loop. But what about the middle branch? Let’s assign a downward current $I_3$ to the middle branch. If we apply Kirchhoff’s junction law $\sum I_{\text{in}}=\sum I_{\text{out}}$ to the junction above the $100\text{Ω}$ resistor, as shown in the blow-up of Figure 28.31, we see that $I_1=I_2+I_3$ and thus $I_3=I_1-I_2$. If $I_3$ ends up being a positive number, then the current in the middle branch really is downward. A negative $I_3$ will signify an upward current.

FIGURE 28.31 Applying Kirchhoff’s laws.

Kirchhoff’s loop law for the left loop, going clockwise from the lower left corner, is

We’re traveling through the $100\text{Ω}$ resistor in the direction of $I_3$, the “downhill” direction, so the potential decreases. The 12 V battery is traversed positive to negative, so there we have $\text{Δ}V=-\epsilon =-12\text{V}$. For the right loop, we’re going to travel “uphill” through the $100\text{Ω}$ resistor, opposite to $I_3$, and gain potential. Thus the loop law for the right loop is

If we substitute $I_3=I_1-I_2$ and then rearrange the terms, we find that the two independent loops have given us two simultaneous equations in the two unknowns $I_1$ and $I_2\text{:}$

We can eliminate $I_2$ by multiplying through the first equation by 3 and then adding the two equations. This gives $1100I_1=33$, from which $I_1=0.030\text{A}=30\text{mA}$. Using this value in either of the two loop equations gives $I_2=0.050\text{A}=50\text{mA}$. Because $I_2>I_1$, the current through the $100\text{Ω}$ resistor is $I_3=I_1-I_2=-20\text{mA,}$ or, because of the minus sign, 20 mA upward. The potential difference across the $100\text{Ω}$ resistor is $\text{Δ}{V}_{100\text{Ω}}=I_{3}R=2.0\text{V}$, with the bottom end more positive.

REVIEW The three “legs” of the circuit are in parallel, so they must have the same potential difference across them. The left leg has $\text{Δ}V=19\text{V}-\left(0.030\text{A}\right)\left(300\text{Ω}\right)=10\text{V}$, the middle leg has $\text{Δ}V=12\text{V}-\left(0.020\text{A}\right)\left(100\text{Ω}\right)=10\text{V}$, and the right leg has $\text{Δ}V=(0.050\text{A})\left(200\text{Ω}\right)=10\text{V}$. Consistency checks such as these are very important. Had we made a numerical error in our circuit analysis, we would have caught it at this point.