Circuit textbooks often write Ohm’s law as $V=IR$ rather than $I=\text{Δ}V\text{/}R$. This can be misleading until you have sufficient experience with circuit analysis. First, Ohm’s law relates the current to the potential difference between the ends of the conductor. Engineers and circuit designers mean “potential difference” when they use the symbol $V$, but the symbol is easily misinterpreted as simply “the potential.” Second, $V=IR$ or even $\text{Δ}V=IR$ suggests that a current $I$ causes a potential difference $\text{Δ}V$. As you have seen, current is a consequence of a potential difference; hence $I=\text{Δ}V\text{/}R$ is a better description of cause and effect.

Despite its name, Ohm’s law is not a law of nature. It is limited to those materials whose resistance $R$ remains constant—or very nearly so—during use. The materials to which Ohm’s law applies are called ohmic. FIGURE 27.20a shows that the current through an ohmic material is directly proportional to the potential difference. Doubling the potential difference doubles the current. Metal and other conductors are ohmic devices.

FIGURE 27.20 Current-versus-potential-difference graphs for ohmic and nonohmic materials.

Full Alternative Text

Because the resistance of metals is small, a circuit made exclusively of metal wires would have enormous currents and would quickly deplete the battery. It is useful to limit the current in a circuit with ohmic devices, called resistors, whose resistance is significantly larger than the metal wires. Resistors are made with poorly conducting materials, such as carbon, or by depositing very thin metal films on an insulating substrate.

Some materials and devices are nonohmic, meaning that the current through the device is not directly proportional to the potential difference. For example, FIGURE 27.20b shows the $I\text{-}\text{versus}\text{-}\text{Δ}V$ graph of a commonly used semiconductor device called a diode. Diodes do not have a well-defined resistance. Batteries, where $\text{Δ}V=\epsilon$ is determined by chemical reactions, and capacitors, where the relationship between $I$ and $\text{Δ}V$ differs from that of a resistor, are important nonohmic devices.

We can identify three important classes of ohmic circuit materials:

1. Wires are metals with very small resistivities $\rho$ and thus very small resistances $(R\ll 1\text{Ω})$. An ideal wire has $R=0\text{Ω}\text{;}$ hence the potential difference between the ends of an ideal wire is $\text{Δ}V=0\text{V}$ even if there is a current in it. We will usually adopt the ideal-wire model of assuming that any connecting wires in a circuit are ideal.

2. Resistors are poor conductors with resistances usually in the range ${10}^1$ to ${10}^6\text{Ω}$. They are used to control the current in a circuit. Most resistors in a circuit have a specified value of $R$, such as $500\text{Ω}$.

3. Insulators are materials such as glass, plastic, or air. An ideal insulator has $R=\infty \text{Ω}\text{;}$ hence there is no current in an insulator even if there is a potential difference across it $(I=\text{Δ}V\text{/}R=0\text{A})$. This is why insulators can be used to hold apart two conductors at different potentials. All practical insulators have $R\gg {10}^9\text{Ω}$ and can be treated, for our purposes, as ideal.

FIGURE 27.21a shows a resistor connected to a battery with current-carrying wires. There are no junctions; hence the current $I$ through the resistor is the same as the current in each wire. Because the wire’s resistance is much less than that of the resistor, $R_{\text{wire}}\ll R_{\text{resist}}$, the potential difference $\text{Δ}{V}_{\text{wire}}=I{R}_{\text{wire}}$ between the ends of each wire is much less than the potential difference $\text{Δ}{V}_{\text{resist}}=I{R}_{\text{resist}}$ across the resistor.

FIGURE 27.21 The potential along a wire-resistor-wire combination.

If we assume ideal wires with $R_{\text{wire}}=0\text{Ω}$, then $\text{Δ}{V}_{\text{wire}}=0\text{V}$ and all the voltage drop occurs across the resistor. In this ideal-wire model, shown in FIGURE 27.21b, the wires are equipotentials, and the segments of the voltage graph corresponding to the wires are horizontal. As we begin circuit analysis in the next chapter, we will assume that all wires are ideal unless stated otherwise. Thus our analysis will be focused on the resistors.

CHAPTER 27 CHALLENGE EXAMPLE Measuring body composition

The woman in the photo is gripping a device that measures body fat. To illustrate how this works, FIGURE 27.22 models an upper arm as part muscle and part fat, showing the resistivities of each. Nonconductive elements, such as skin and bone, have been ignored. This is obviously not a picture of the actual structure, but gathering all the fat tissue together and all the muscle tissue together is a model that predicts the arm’s electrical character quite well.

FIGURE 27.22 A simple model for the resistance of an arm.

FIGURE 27.22 Full Alternative Text

A 0.87 mA current is recorded when a 0.60 V potential difference is applied across an upper arm having the dimensions shown in the figure. What are the percentages of muscle and fat in this person’s upper arm?

MODEL Model the muscle and the fat as separate resistors connected to a 0.60 V battery. Assume the connecting wires to be ideal, with no “loss” of potential along the wires.

VISUALIZE FIGURE 27.23 shows the circuit, with the side-by-side muscle and fat resistors connected to the two terminals of the battery.

FIGURE 27.23 Circuit for passing current through the upper arm.

FIGURE 27.23 Full Alternative Text

SOLVE The measured current of 0.87 mA is $I_{\text{total}}$, the current traveling from the battery to the arm and later back to the battery. This current splits at the junction between the two resistors. Kirchhoff’s junction law, for the conservation of charge, requires

$$I_{\text{total}}=I_{\text{muscle}}+I_{\text{fat}}$$

The current through each resistor can be found from Ohm’s law: $I=\text{Δ}V\text{/}R$. Each resistor has $\text{Δ}V=0.60\text{V}$ because each is connected to the battery terminals by lossless, ideal wires, but they have different resistances.

Let the fraction of muscle tissue be $x\text{;}$ the fraction of fat is then $1-x$. If the cross-section area of the upper arm is $A=\pi r^2$, then the muscle resistor has $A_{\text{muscle}}=xA$ while the fat resistor has $A_{\text{fat}}=(1-x)A$. The resistances are related to the resistivities and the geometry by

$$\begin{array}{l}{R}_{\text{muscle}}={\displaystyle \frac{{\rho }_{\text{muscle}}L}{A_{\text{muscle}}}}={\displaystyle \frac{{\rho }_{\text{muscle}}L}{x\pi r^2}}\hfill \\ R_{\text{fat}}={\displaystyle \frac{{\rho }_{\text{fat}}L}{A_{\text{fat}}}}={\displaystyle \frac{{\rho }_{\text{fat}}L}{\left(1-x\right)\pi r^2}}\hfill \end{array}$$

The currents are thus

$$\begin{array}{l}{I}_{\text{muscle}}={\displaystyle \frac{\text{Δ}V}{R_{\text{muscle}}}}={\displaystyle \frac{x\pi r^2\text{Δ}V}{{\rho }_{\text{muscle}}L}}=0.93x\text{ mA}\hfill \\ I_{\text{fat}}={\displaystyle \frac{\text{Δ}V}{R_{\text{fat}}}}={\displaystyle \frac{\left(1-x\right)\pi r^2\text{Δ}V}{{\rho }_{\text{fat}}L}}=0.48\left(1-x\right)\text{ mA}\hfill \end{array}$$

The sum of these is the total current:

$$\begin{array}{ccc}\hfill I_{\text{total}}& \hfill =\hfill & 0.87\text{mA}=0.93x\text{mA}+0.48(1-x)\text{mA}\hfill \\ \hfill & \hfill =\hfill & (0.48+0.45x)\text{mA}\hfill \end{array}$$

Solving, we find $x=0.87$. This subject’s upper arm is 87% muscle tissue, 13% fat tissue.

REVIEW The percentages seem reasonable for a healthy adult. A real measurement of body fat requires a more detailed model of the human body, because the current passes through both arms and across the chest, but the principles are the same.