Using thick connecting wires that are very good conductors, a Nichrome wire (“wire 1”) of length L₁ and cross-sectional area A₁ is connected in series with a battery and an ammeter (this is circuit 1). The reading on the ammeter is I₁. Now the Nichrome wire is removed and replaced with a different wire (“wire 2”), which is 2.5 times as long and has 5.5 times the cross-sectional area of the original wire (this is circuit 2). In the following question, a subscript 1 refers to circuit 1, and a subscript 2 refers to circuit 2. It will be helpful to write out your solutions to the following questions algebraically before doing numerical calculations. (Hint: Think about what is the same in these two circuits.)(a) What is the value of I₂/ I₁, the ratio of the conventional currents in the two circuits? (b) What is the value of R₂/ R₁, the ratio of the resistances of the wires? (c) What is the value of E₂/ E₁, the ratio of the electric fields inside the wires in the steady states?

In circuit 1, a Nichrome wire with a length of $L_1$and cross-sectional area of$A_1$ is connected in series with a battery and an ammeter to measure the current, and for circuit 2, the Nichrome wire is replaced with a different wire with a length which is 2.5 times as long and the crossectional area is 5.5 times the cross-sectional area of the original wire. That means that now the length is$L_2=2.5L_1$ and the cross-sectional area is$A_2=5.5A_1$ .

The electric field generated by current flowing in a wire can be written mathematically as,

$\mathit{E}\mathbf{=}\frac{\mathbf{V}}{\mathbf{L}}$ (1)

Here V is the applied voltage in the circuit, L is the length of the wire.

For circuit 1, use equation 1 to calculate the electric field inside the wire as,

$E_1=\frac{V}{L_1}$ (2)

For circuit 2, use equation 1 to calculate the electric field inside the wire as,

$E_2=\frac{V}{L_2}$ (3)

Divide equation 3 by equation 2 as,

$$\begin{gathered} \frac{E_2}{E_1}=\frac{\frac{V}{L_2}}{\frac{V}{L_1}} \\ \frac{E_2}{E_1}=\frac{\frac{1}{L_2}}{\frac{1}{L_1}} \end{gathered}$$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}\frac{E_2}{E_1}& =& \frac{\frac{1}{2.5L_1}}{\frac{1}{L_1}}\\ & =& \frac{1}{2.5}\\ \frac{E_2}{E_1}& =& 0.4\end{array}$

Thus, the ratio of the electric field inside the wires is $\frac{E_2}{E_1}=0.4$.