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?

For circuit 1,

A Nichrome wire (“wire 1”) of length $L_1$and cross-sectional area $A_1$is connected in series with a battery and an ammeter.

For circuit 2,

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, which means the length now is $L_2=2.5L_1$and the cross-sectional area is $A_2=5.5A_1$.

The current flowing in a circuit can be written mathematically as,

$\mathit{I}\mathbf{=}\frac{\mathbf{V}\mathbf{A}}{\mathbf{\rho }\mathbf{L}}$ (1)

Here V is the applied voltage in the circuit; A is the cross-sectional area of the connecting wire that is used in the circuit; L is the length of the wire, role="math" localid="1663003300566" $\mathit{\rho }\mathbf{}$is the resistivity of the material out of which the wire was made.

For circuit 1, use equation 1 to calculate the flowing current as,

$I_1=\frac{V{A}_1}{\rho L_1}$ (2)

For circuit 2, use equation 1 to calculate the flowing current as,

$I_2=\frac{V{A}_2}{\rho L_2}$ (3)

Divide equation 3 by equation 2 as,

$\begin{array}{rcl}\frac{I_2}{I_1}& =& \frac{\frac{V{A}_2}{\rho L_2}}{\frac{V{A}_1}{\rho L_1}}\\ \frac{I_2}{I_1}& =& \frac{\frac{A_2}{L_2}}{\frac{A_1}{L_1}}\end{array}$

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

$\begin{array}{rcl}\frac{I_2}{I_1}& =& \frac{\frac{5.5A_1}{2.5L_1}}{\frac{A_1}{L_1}}\\ & =& \frac{5.5}{2.5}\\ \frac{I_2}{I_1}& =& 2.2\\ & & \end{array}$

Thus, the ratio of conventional currents in the two circuits is$\frac{I_2}{I_1}=2.2$ .