Chapter 19: Q60P-b (page 800)

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?

Short Answer

Expert verified

(b) The ratio of the resistances of two wires is $\frac{R_{2}}{R_{1}} = 0.4545$ .

Step by step solution

Given data

For circuit 1, the 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, and for circuit 2, the Nichrome wire is removed and 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 $L_{2} = 2.5 L_{1}$ and the cross-sectional area is $A_{2} = 5.5 A_{1}$ .

Concept

The resistance of the wires used in a circuit can be measured as,

$R = \frac{ρ L}{A}$ (1)

Here A is the cross-sectional area of the connecting wire that is used in the circuit, L is the length of the wire, $ρ$ is the resistivity of the material out of which the wire was made.

(b) Calculate the ratio of the resistances of the wires

For circuit 1, use equation 1 to calculate the resistance as,

$R_{1} = \frac{ρ L_{1}}{A_{1}}$ (2)

For circuit 2, use equation 1 to calculate the resistance as,

$R_{2} = \frac{ρ L_{2}}{A_{2}}$ (3)

Divide equation 3 by equation 2 as,

$\begin{array}{rcl} \frac{R_{2}}{R_{1}} & = & \frac{\frac{ρ L_{2}}{A_{2}}}{\frac{ρ L_{1}}{A_{1}}} \\ \frac{R_{2}}{R_{1}} & = & \frac{\frac{L_{2}}{A_{2}}}{\frac{L_{1}}{A_{1}}} \end{array}$

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

$\begin{array}{rcl} \frac{R_{2}}{R_{1}} & = & \frac{\frac{2.5 L_{1}}{5.5 A_{1}}}{\frac{L_{1}}{A_{1}}} \\ = & \frac{2.5}{5.5} \\ \frac{R_{2}}{R_{1}} & = & 0.4545 \end{array}$