28.2.1 The Basic Circuit

The most basic electric circuit is a single resistor connected to the two terminals of a battery. FIGURE 28.6a shows a literal picture of the circuit elements and the connecting wires; FIGURE 28.6b is the circuit diagram. Notice that this is a complete circuit, forming a continuous path between the battery terminals.

The figure labeled (“a”) illustrates a cartoon of a resistor connected to a battery.

The figure labeled (b) illustrates the circuit diagram of a resistor connected to a battery.

Full Alternative Text

The resistor might be a known resistor, such as “a $10 Ω$ resistor,” or it might be some other resistive device, such as a lightbulb. Regardless of what the resistor is, it is called the load. The battery is called the source.

FIGURE 28.7 shows the use of Kirchhoff’s loop law to analyze this circuit. Two things are worth noting:

This circuit has no junctions, so the current $I$ is the same in all four sides of the circuit. Kirchhoff’s junction law is not needed.
We’re assuming the ideal-wire model, in which there are no potential differences along the connecting wires.

The figure illustrates an analysis of the basic circuit using Kirchhoff’s loop law.

FIGURE 28.7 Full Alternative Text

Kirchhoff’s loop law, with two circuit elements, is

\begin{array}{l} Δ V_{loop} & = \sum {(Δ V)}_{i} = Δ V_{bat} + Δ V_{res} \\ = 0 \end{array}

(28.3)

Let’s look at each of the two voltages in Equation 28.3:

The potential increases as we travel through the battery on our clockwise journey around the loop. We enter the negative terminal and, farther downstream, exit the positive terminal after having gained potential $ε$ . Thus

$Δ V_{bat} = + ε$
The potential of a conductor decreases in the direction of the current, which we’ve indicated with the $+$ and $-$ signs in Figure 28.7. Thus

$Δ V_{res} = V_{downstream} - V_{upstream} = - I R$

NOTE Determining which potential differences are positive and which are negative is perhaps the most important step in circuit analysis.

With this information, the loop equation becomes

ε - I R = 0

(28.4)

We can solve the loop equation to find that the current in the circuit is

I = \frac{ε}{R}

(28.5)

We can then use the current to find that the magnitude of the resistor’s potential difference is

Δ V_{R} = I R = ε

(28.6)

This result should come as no surprise. The potential energy that the charges gain in the battery is subsequently lost as they “fall” through the resistor.

NOTE The current that the battery delivers depends jointly on the emf of the battery and the resistance of the load.

EXAMPLE 28.1 Two resistors and two batteries

Analyze the circuit shown in FIGURE 28.8.

a. Find the current in and the potential difference across each resistor.

b. Draw a graph showing how the potential changes around the circuit, starting from $V = 0 V$ at the negative terminal of the 6 V battery.

The figure illustrates a circuit diagram consisting of two batteries and two resistors.

FIGURE 28.8 Full Alternative Text

MODEL Assume ideal connecting wires and ideal batteries, for which $Δ V_{bat} = ε$ .

VISUALIZE In FIGURE 28.9, we’ve redrawn the circuit and defined $ε_{1}, ε_{2}, R_{1}$ , and $R_{2}$ . Because there are no junctions, the current is the same through each component in the circuit. With some thought, we might deduce whether the current is cw or ccw, but we do not need to know in advance of our analysis. We will choose a clockwise direction and solve for the value of I. If our solution is positive, then the current really is cw. If the solution should turn out to be negative, we will know that the current is ccw.

The figure illustrates the analysis of a circuit diagram consisting of two batteries and two resistors.

FIGURE 28.9 Full Alternative Text

SOLVE

a. How do we deal with two batteries? Can charge flow “backward” through a battery, from positive to negative? Consider the charge escalator analogy. Left to itself, a charge escalator lifts charge from lower to higher potential. But it is possible to run down an up escalator, as many of you have probably done. If two escalators are placed “head to head,” whichever is stronger will, indeed, force the charge to run down the up escalator of the other battery. The current in a battery can be from positive to negative if driven in that direction by a larger emf from a second battery. Indeed, this is how rechargeable batteries are recharged.

Kirchhoff’s loop law, going clockwise from the negative terminal of battery 1, is

\begin{array}{l} Δ V_{loop} & = \sum Δ V)_{i} \\ = Δ V_{bat 1} + Δ V_{res 1} + Δ V_{bat 2} + Δ V_{res 2} = 0 \end{array}

All the signs are $+$ because this is a formal statement of adding potential differences around the loop. Next we can evaluate each $Δ V$ . As we go cw, the charges gain potential in battery 1 but lose potential in battery 2. Thus $Δ V_{bat 1} = + ε_{1}$ and $Δ V_{bat 2} = - ε_{2}$ . There is a loss of potential in traveling through each resistor, because we’re traversing them in the direction we assigned to the current, so $Δ V_{res 1} = - I R_{1}$ and $Δ V_{res 2} = - I R_{2}$ . Thus Kirchhoff’s loop law becomes

\begin{matrix} \sum {(Δ V)}_{i} & = & ε_{1} - I R_{1} - ε_{2} - I R_{2} \\ = & ε_{1} - ε_{2} - I (R_{1} + R_{2}) = 0 \end{matrix}

We can solve this equation to find the current in the loop:

I = \frac{ε_{1} - ε_{2}}{R_{1} + R_{2}} = \frac{6 V - 9 V}{4 Ω + 2 Ω} = - 0.50 A

The value of $I$ is negative; hence the actual current in this circuit is 0.50 A counterclockwise. You perhaps anticipated this from the orientation of the 9 V battery with its larger emf.

The potential difference across the $4 Ω$ resistor is

Δ V_{res 1} = - I R_{1} = - (- 0.50 A) (4 Ω) = + 2.0 V

Because the current is actually ccw, the resistor’s potential increases in the cw direction of our travel around the loop. Similarly, $Δ V_{res 2} = 1.0 V$ .

b. FIGURE 28.10 shows the potential experienced by charges flowing around the circuit. The distance $s$ is measured from the 6 V battery’s negative terminal, and we have chosen to let $V = 0 V$ at that point. The potential ends at the value from which it started.

The figure illustrates the potential versus distance graph.

FIGURE 28.10 Full Alternative Text

REVIEW Notice how the potential drops 9 V upon passing through battery 2 in the cw direction. It then gains 1 V upon passing through $R_{2}$ to end at the starting potential.

28.2.1 The Basic Circuit

Figure 28.6 The basic circuit of a resistor connected to a battery.

FIGURE 28.7 Analysis of the basic circuit using Kirchhoff’s loop law.

EXAMPLE 28.1 Two resistors and two batteries

FIGURE 28.8 Circuit for Example 28.1.

FIGURE 28.9 Analyzing the circuit.

FIGURE 28.10 A graphical presentation of how the potential changes around the loop.