Question

In the circuit shown in Figure 18.87, bulbs 1 and 2 are identical in mechanical construction (the filaments have the same length and the same cross-sectional area), but the filaments are made of different metals. The electron mobility in the metal used in bulb 2 is three times as large as the electron mobility in the metal used in bulb 1, but both metals have the same number of mobile electrons per cubic meter. The two bulbs are connected in series to two batteries with thick copper wires (like your connecting wires).

(a)In bulb 1, the electron current is $i_1$and the electric field is $E_1$. In terms of these quantities, determine the corresponding quantities $i_2$and $E_2$for bulb 2, and explain your reasoning.

(b)When bulb 2 is replaced by a wire, the electron current through bulb 1 is $i_0$and the electric field in bulb 1 is $E_0$. How big is $i_1$ in terms of $i_0$? Explain your answer, including explicit mention of any approximations you must make. Do not use ohms or series-resistance equations in your explanation, unless you can show in detail how these concepts follow from the microscopic analysis introduced in this chapter.

(c)Explain why the electric field inside the thick copper wires is very small. Also explain why this very small electric field is the same in all of the copper wires, if they all have the same cross-sectional area.

(d)Figure 18.88 is a graph of the magnitude of the electric field at each location around the circuit when bulb 2 is replaced by a wire. Copy this graph and add to it, on the same scale, a graph of the magnitude of the electric field at each location around the circuit when both bulbs are in the circuit. The very small field in the copper wires has been shown much larger than it really is in order to give you room to show how that small field differs in the two circuits.

Short answer

(a) The current in the two bulbs is related as i₁= i₂ and the field in the two bulbs is related as $E_2=\frac{E_1}{3}$.

Step-by-step solution

Given data

Electron mobility in bulb 2 is related to electron mobility in bulb 1 as

u₂ = 3u₁ .................(1)

The wires in the two bulbs have the same length, cross sectional area and charge density.

Electron current in a circuit

Electron current in a wire of free charge density n , cross sectional area A , mobility u and electric field E is:

i = nAuE .............(2)

(a) Comparison of the currents and electric fields in the two bulbs

Since the two bulbs are in series, the current in them are the same, that is,

i₁ = i₂ .......(3)

Both the bulbs have the same charge density and cross sectional area. Thus, from equation (1), the first bulb has current

i₁ = nAu₁E .................(4)

and the second bulb has current

i₂ = nAu₂E .................(5)

Divide equation (5) by equation (iv) and use equations (3) and (1) to get:

$$\begin{gathered} \frac{i_2}{i_1}=\frac{nA{u}_2{E}_2}{nA{u}_1{E}_1} \\ 1=\frac{3u_1{E}_2}{u_1{E}_1} \\ {E}_2=\frac{E_1}{3} \end{gathered}$$

Thus, we have i₁ = i₂ and $E_2=\frac{E_1}{3}$.