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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Two different electrical devices have the same power consumption, but one is meant to be operated on $$120 - V$$ AC and the other on $$240 - V$$AC. (a) What is the ratio of their resistances? (b) What is the ratio of their currents? (c) Assuming its resistance is unaffected, by what factor will the power increase if a $$120 - V$$ AC device is connected to $$240 - V$$ AC?

1. Ratio of the resistance is $$\frac{{{R_2}}}{{{R_1}}} = 4.0$$.
2. Ratio of the current is $$\frac{{{I_2}}}{{{I_1}}} = 0.50$$.
3. Ratio of the powers is $$\frac{{{P_2}}}{{{P_1}}} = 4.0$$.
See the step by step solution

## Step 1: Relation between power and potential difference.

The power, or rate at which energy is delivered to a circuit element, is $$P = I\Delta V$$ if a potential difference $$\Delta V$$ is maintained across the element.

We can represent the power applied to a resistor as \begin{align}{}P = {I^2}R\\P = \frac{{{{(\Delta V)}^2}}}{R}\end{align} since the potential difference across a resistor is given by $$\Delta V = IR$$.

A resistor's inherent energy manifests as the energy transmitted to it through electrical transmission.

## Step 2: Given data

• The voltage across the first device is: $$\Delta {V_1} = 120\;V$$.
• The voltage across the second device is: $$\Delta {V_2} = 240\;V$$.
• Both devices have the same power consumption: $${P_1} = {P_2}$$.

## Step 3: Calculation of resistance ratio.

a.

The power consumption of the first device is found from equation:

$${P_1} = \frac{{{{\left( {\Delta {V_1}} \right)}^2}}}{{{R_1}}}$$

Similarly, the power consumption of the second device is:

$${P_2} = \frac{{{{\left( {\Delta {V_2}} \right)}^2}}}{{{R_2}}}$$

But the two devices have the same power consumption.

So $${P_1} = {P_2}$$.

$$\frac{{{{\left( {\Delta {V_1}} \right)}^2}}}{{{R_1}}} = \frac{{{{\left( {\Delta {V_2}} \right)}^2}}}{{{R_2}}}$$

Solve for $${R_2}/{R_1}$$:

$$\frac{{{R_2}}}{{{R_1}}} = \frac{{{{\left( {\Delta {V_2}} \right)}^2}}}{{{{\left( {\Delta {V_1}} \right)}^2}}}$$

Entering the values for $$\Delta {V_2}$$and $$\Delta {V_1}$$we obtain:

\begin{align}{}\frac{{{R_2}}}{{{R_1}}} = {\left( {\frac{{240\;V}}{{120\;V}}} \right)^2}\\\frac{{{R_2}}}{{{R_1}}} = 4.0\end{align}

Therefore, the resistance ratio is $$4.0$$.

## Step 4: Calculation of the ratio of current.

(b)

The power consumption of the first device is found in terms of the current and the resistance by equation $$(1)$$:

$${P_1} = I_1^2{R_1}$$

Similarly, the power consumption of the second device is:

$${P_2} = I_2^2{R_2}$$

Again, we use the fact that both powers are equal:

\begin{align}{}{P_1} = {P_2}\\I_1^2{R_1} = I_2^2{R_2}\end{align}

Solve for $${I_2}/{I_1}$$:

$$\frac{{{I_2}}}{{{I_1}}} = \sqrt {\frac{{{R_1}}}{{{R_2}}}}$$

Substitute for $${R_1}/{R_2}$$ from part (a):

\begin{align}{}\frac{{{I_2}}}{{{I_1}}} = \sqrt {\frac{1}{{4.0}}} \\\frac{{{I_2}}}{{{I_1}}} = 0.50\end{align}

Therefore, the ratio of the current is $$0.50$$.

## Step 5: Calculation of power ratio.

c.

According to Equation the power consumed by the device is directly proportional to the voltage squared provided that the resistance is constant.

So

$$P \propto {(\Delta V)^2}$$

Expressing this proportionality as two equal fractions, we obtain:

\begin{align}{}\frac{{{P_2}}}{{{P_1}}} = {\left( {\frac{{\Delta {V_2}}}{{\Delta {V_1}}}} \right)^2}\\\frac{{{P_2}}}{{{P_1}}} = {\left( {\frac{{240\;V}}{{120\;V}}} \right)^2}\\\frac{{{P_2}}}{{{P_1}}} = 4.0\end{align}

Hence, the power ratio is $$4.0$$.