Question

The picture shows a battery connected to two wires in parallel. Both wires are made of the same material and are of the same length, but the diameter of wire A is twice the diameter of wire B.

True or False for the following:

1) The resistance of wire B is four times as large as the resistance of wire A.

2) The power dissipated in wire A is 16 times the power dissipated in wire B.

3) The voltage drop across wire B is larger than the voltage drop across wire A.

4) The resistance of wire B is twice as large as the resistance of wire A.

5) The current through the battery is five times larger than the current through wire B.

(please explain! thanks!)

Answer 1

Concepts and reason

The concepts used here are the relation of resistance and resistivity of a material, electric power and the Ohm’s law. The given statements can be analyzed using the above mentioned concepts.

Fundamentals

Resistance and Resistivity:

The relation between resistance and resistivity is:

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

Here, $\rho$ is the resistivity, $R$ is the resistance, $L$ is the length and $A$ is cross-sectional area.

Ohm’s Law:

According to Ohm’s Law, “the voltage is directly proportional to the current”

$\begin{array}{l}\\V = IR\\\\I = \frac{V}{R}\\\end{array}$

Here, $V$ is the voltage, $I$ is the current and $R$ is the resistance.

Electric power:

The electric power in terms of voltage and resistance is:

$P = \frac{{{V^2}}}{R}$

Here, $V$ is the voltage and $R$ is the resistance.

(1)

The relation between resistance and resistivity is:

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

Area of wire is: $A = \pi {r^2}$

Here, $r$ is the radius of wire.

$\Rightarrow R = \frac{{\rho L}}{{\pi {r^2}}}$

For wire A, the resistance is:

${R_{\rm{A}}} = \frac{{\rho L}}{{\pi {r_{\rm{A}}}^2}}$

Here, ${r_{\rm{A}}}$ is the radius of wire A.

For wire B, the resistance is:

${R_{\rm{B}}} = \frac{{\rho L}}{{\pi {r_{\rm{B}}}^2}}$

Here, ${r_{\rm{B}}}$ is the radius of wire B.

Substitute ${r_{\rm{A}}} = 2{r_{\rm{B}}}$ .

$\begin{array}{c}\\{R_{\rm{A}}} = \frac{{\rho L}}{{\pi {{\left( {2{r_{\rm{B}}}} \right)}^2}}}\\\\ = \frac{{\rho L}}{{4\pi {r_{\rm{B}}}^2}}\\\\ = \frac{1}{4}\left( {\frac{{\rho L}}{{\pi {r_{\rm{B}}}^2}}} \right)\\\end{array}$

Substitute $\frac{{\rho L}}{{4\pi {r_{\rm{B}}}^2}}$ as ${R_{\rm{B}}}$ .

$\begin{array}{l}\\{R_{\rm{A}}} = \frac{1}{4}\left( {{R_{\rm{B}}}} \right)\\\\{R_{\rm{B}}} = 4{R_{\rm{A}}}\\\end{array}$

The power dissipated in wire A is:

${P_{\rm{A}}} = \frac{{{V^2}}}{{{R_{\rm{A}}}}}$

The power dissipated in wire B is:

${P_{\rm{B}}} = \frac{{{V^2}}}{{{R_{\rm{B}}}}}$

The wires are connected in parallel, so the voltage across them is equal.

Substitute $4{R_{\rm{A}}}$ for ${R_{\rm{B}}}$ .

$\begin{array}{c}\\{P_{\rm{B}}} = \frac{{{V^2}}}{{4{R_{\rm{A}}}}}\\\\ = \frac{1}{4}\left( {\frac{{{V^2}}}{{{R_{\rm{A}}}}}} \right)\\\end{array}$

Substitute $\frac{{{V^2}}}{{{R_{\rm{A}}}}}$ as ${P_{\rm{A}}}$ .

$\begin{array}{l}\\{P_{\rm{B}}} = \frac{1}{4}\left( {{P_{\rm{A}}}} \right)\\\\{P_{\rm{A}}} = 4{P_{\rm{B}}}\\\end{array}$

The power dissipated in wire A is four times the power dissipated in wire B.

It is given that the wires are connected in parallel, so the voltage across both the wires is equal.

The resistance of wire b is four times the resistance of wire A as proved above.

${R_{\rm{B}}} = 4{R_{\rm{A}}}$

Let the current through the battery be $I$ , current through wire A be ${I_{\rm{A}}}$ and current through wire B be ${I_{\rm{B}}}$ .

As the wires are connected in parallel, the current through the battery should be equal to the sum of the current flowing through the wires.

$I = {I_{\rm{A}}} + {I_{\rm{B}}}$

Substitute $\frac{V}{{{R_{\rm{A}}}}}$ for ${I_{\rm{A}}}$ and $\frac{V}{{{R_{\rm{B}}}}}$ for ${I_{\rm{B}}}$ .

$I = \frac{V}{{{R_{\rm{A}}}}} + \frac{V}{{{R_{\rm{B}}}}}$

Substitute $\frac{{{R_{\rm{B}}}}}{4}$ for ${R_{\rm{A}}}$ .

$\begin{array}{c}\\I = \frac{V}{{\frac{{{R_{\rm{B}}}}}{4}}} + \frac{V}{{{R_{\rm{B}}}}}\\\\ = \frac{{4V}}{{{R_{\rm{B}}}}} + \frac{V}{{{R_{\rm{B}}}}}\\\\ = \frac{{4V + V}}{{{R_{\rm{B}}}}}\\\\ = 5\frac{V}{{{R_{\rm{B}}}}}\\\end{array}$

Substitute $\frac{V}{{{R_{\rm{B}}}}}$ for ${I_{\rm{B}}}$ .

$\begin{array}{c}\\I = 5\left( {\frac{V}{{{R_{\rm{B}}}}}} \right)\\\\ = 5{I_{\rm{B}}}\\\end{array}$

The current through the battery is five times more than the current through wire B.

Ans: Part 1

The statement 1 is “True”.

Part 2

The statement 2 is “False”.

Part 3

The statement 3 is “False”.

Part 4

The statement 4 is “False”.

Part 5

The statement 5 is “True”.