Chapter 7, Problem 35Q

(Indicates a review question, which means it requires only a basic understanding of the material to answer. Questions without this designation typically require integrating or extending the concepts presented thus far.)

. Six resistors, all having the same temperatures and compositions, are each attached to a battery with voltage V to form a simple circuit. The resistors have different lengths, l, but the same diameters. Rank the circuits according to the current that flows through the resistor from smallest to greatest. If two (or more) circuits have the same current flowing, give them the same ranking. Explain the rationale you used to establish your rankings.
Circuit 1: l = 2 cm; V = 3 V
Circuit 2: l = 2 cm; V = 6 V
Circuit 3: l = 1 cm; V = 3 V
Circuit 4: l = 1 cm; V = 6 V
Circuit 5: l = 3cm; V = 2 V
Circuit 6: l = 3cm; V = 6 V

To determine

The ranking from smallest to greatest of the given circuits in terms of currents carried by them.

Answer to Problem 35Q

Ranking of the circuits according to the current that flows through the resistor from smallest to greatest are as follows:

Rank 1: Circuit 5: $I_5=0.67\frac{A}{\rho }\text{ampere}$

Rank 2: Circuit 1: $I_1=1.5\frac{A}{\rho }\text{ampere}$

Rank 3: Circuit 6: $I_6=2\frac{A}{\rho }\text{ampere}$

Rank 4: Circuit 2: $I_2=3\frac{A}{\rho }\text{ampere}$

Rank 4: Circuit 3: $I_3=3\frac{A}{\rho }\text{ampere}$

Rank 5: Circuit 4: $I_4=6\frac{A}{\rho }\text{ampere}$.

Explanation of Solution

Given information:

All the resistors have same composition, temperature and diameter. And each circuit comprises simple connection of resistor and the battery having voltage (V). The parameters for the given circuits are as follows:

Circuit 1: $l=2\text{cm;}\text{V}=3\text{V}$

Circuit 2: $l=2\text{cm;}\text{V}=6\text{V}$

Circuit 3: $l=1\text{cm;}\text{V}=3\text{V}$

Circuit 4: $l=1\text{cm;}\text{V}=6\text{V}$

Circuit 5: $l=3\text{cm;}\text{V}=2\text{V}$

Circuit 6: $l=3\text{cm;}\text{V}=6\text{V}$

Concept Used:

1. Ohm’s law states that

$V=IR$

2. The resistance $\left(R\right)$ of a wire having resistivity, area and length as $\rho$ , $A$ and $l$ respectively, is as:

   $R=\rho \frac{l}{A}$

   Where, $\rho$ is the resistivity of the conductor.

   l is the length of the conductor and

   A is the area of the conductor.

3. Resistivity of a resistor depends upon its composition and temperature.

Calculation:

Since, all the resistors have same temperature and composition therefore, the value of resistivity will be same for them. And the diameters of all the resistors are same which implies that the area of these resistor will be same.

Let R₁, R₂, R₃, R₄, R₅ & R₆ be the resistances of the circuits 1, 2, 3, 4, 5 & 6 respectively.

Also, let I₁, I₂, I₃, I₄, I₅ & I₆ be the currents flowing through the circuits 1, 2, 3, 4, 5 & 6 respectively.

Therefore, we obtain the parameters for the circuits as follows:

For circuit 1,

$\begin{array}{l}{R}_1=\rho \frac{l}{A}\\ \Rightarrow R_1=2\frac{\rho }{A}\because l=2\text{cm}\end{array}$

Therefore,

$\begin{array}{l}{I}_1=\frac{V}{R_1}\because \text{By}\text{ohm's}\text{law}\\ \Rightarrow I_1=\frac{3}{2\frac{\rho }{A}}\because R_1=2\frac{\rho }{A},\text{V}=3\text{V}\\ \Rightarrow I_1=1.5\frac{A}{\rho }\text{ampere}\end{array}$

For circuit 2,

$\begin{array}{l}{R}_2=\rho \frac{l}{A}\\ \Rightarrow R_2=2\frac{\rho }{A}\because l=2\text{cm}\end{array}$

Therefore,

$\begin{array}{l}{I}_2=\frac{V}{R_2}\because \text{By}\text{ohm's}\text{law}\\ \Rightarrow I_2=\frac{6}{2\frac{\rho }{A}}\because R_2=2\frac{\rho }{A},\text{V}=6\text{V}\\ \Rightarrow I_2=3\frac{A}{\rho }\text{ampere}\end{array}$

For circuit 3,

$\begin{array}{l}{R}_3=\rho \frac{l}{A}\\ \Rightarrow R_3=\frac{\rho }{A}\because l=1\text{cm}\end{array}$

Therefore,

$\begin{array}{l}{I}_3=\frac{V}{R_3}\because \text{By}\text{ohm's}\text{law}\\ \Rightarrow I_3=\frac{3}{\frac{\rho }{A}}\because R_3=\frac{\rho }{A},\text{V}=3\text{V}\\ \Rightarrow I_3=3\frac{A}{\rho }\text{ampere}\end{array}$

For circuit 4,

$\begin{array}{l}{R}_4=\rho \frac{l}{A}\\ \Rightarrow R_4=\frac{\rho }{A}\because l=1\text{cm}\end{array}$

Therefore,

$\begin{array}{l}{I}_4=\frac{V}{R_4}\because \text{By}\text{ohm's}\text{law}\\ \Rightarrow I_4=\frac{6}{\frac{\rho }{A}}\because R_4=\frac{\rho }{A},\text{V}=6\text{V}\\ \Rightarrow I_4=6\frac{A}{\rho }\text{ampere}\end{array}$

For circuit 5,

$\begin{array}{l}{R}_5=\rho \frac{l}{A}\\ \Rightarrow R_5=3\frac{\rho }{A}\because l=3\text{cm}\end{array}$

Therefore,

$\begin{array}{l}{I}_5=\frac{V}{R_5}\because \text{By}\text{ohm's}\text{law}\\ \Rightarrow I_5=\frac{2}{3\frac{\rho }{A}}\because R_5=3\frac{\rho }{A},\text{V}=2\text{V}\\ \Rightarrow I_5=0.67\frac{A}{\rho }\text{ampere}\end{array}$

For circuit 6,

$\begin{array}{l}{R}_6=\rho \frac{l}{A}\\ \Rightarrow R_6=3\frac{\rho }{A}\because l=3\text{cm}\end{array}$

Therefore,

$\begin{array}{l}{I}_6=\frac{V}{R_6}\because \text{By}\text{ohm's}\text{law}\\ \Rightarrow I_6=\frac{6}{3\frac{\rho }{A}}\because R_6=3\frac{\rho }{A},\text{V}=6\text{V}\\ \Rightarrow I_6=2\frac{A}{\rho }\text{ampere}\end{array}$

Thus, we observe that the currents in the ascending order of the given circuits are as:

$I_5<I_1<I_6<I_2=I_3<I_4$.

Conclusion:

Ranking of the circuits according to the current that flows through the resistor from smallest to greatest are as follows:

Rank 1: Circuit 5: $I_5=0.67\frac{A}{\rho }\text{ampere}$

Rank 2: Circuit 1: $I_1=1.5\frac{A}{\rho }\text{ampere}$

Rank 3: Circuit 6: $I_6=2\frac{A}{\rho }\text{ampere}$

Rank 4: Circuit 2: $I_2=3\frac{A}{\rho }\text{ampere}$

Rank 4: Circuit 3: $I_3=3\frac{A}{\rho }\text{ampere}$

Rank 5: Circuit 4: $I_4=6\frac{A}{\rho }\text{ampere}$.