Suppose you have a simple circuit whose wire changes in thickness. The wire is 8 meters long. The first 2 meters of the wire are 3 mm thick. The next 2 meters are 1 mm thick. The last 4 meters are 3 mm thick. The wire is connected to a 12-Volt battery and current is allowed to flow. You use an ammeter and a voltmeter to find that the current through the first 2 meters of wire is $I_1 = 5 \text{ A}$, and the voltage across the first two meters is $\Delta V_1 = 1 \text{ V}$. In all three segments of the wire, determine the magnitude of the electric field inside and the power transmitted.

• Segment lengths: $L_1=2 \text{ m}$, $L_2=2 \text{ m}$, and $L_3=4 \text{ m}$.
• Segment diameters: $d_1=3 \text{ mm}$, $d_2=1 \text{ mm}$, and $d_3=3 \text{ mm}$.
• Current: $I_1 = 5 \text{ A}$.
• Voltage: $\Delta V_1 = 1 \text{ V}$, $\Delta V_{battery} = 12 \text{ V}$.

• Power and electric field in all segments

• The circuit is in a steady state.
• Approximating the battery as a mechanical battery.
• The wire has a circular cross-section.
• No outside influence on the circuit.
• The wire is made of the same material throughout.

• We represent the situation with diagram below. We number the segments for simplicity of representing the quantities we are interested in (see above in “Facts”).

Let's start with segment 1. The electric field is constant since the wire is uniform with respect to the rest of the segment, so we get $$E_1 = \frac{\Delta V_1}{L_1} = 0.5 \text{ V/m}$$ The power dissipated through the segment is just $$P_1=I_1 \Delta V_1 = 5 \text{ W}$$

Now, for segment 2. We can use what we know about charge in steady state circuits to determine the electric field (notice we divide diameter by 2 in order to use the equation for the area of the circular cross-section): $$E_2=\frac{A_1}{A_2}E_1 = \frac{\pi (d_1/2)^2}{\pi (d_2/2)^2}E_1=9E_1=4.5\text{ V/m}$$