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Q81PE

Expert-verifiedFound in: Page 736

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?**

- Ratio of the resistance is \(\frac{{{R_2}}}{{{R_1}}} = 4.0\).
- Ratio of the current is \(\frac{{{I_2}}}{{{I_1}}} = 0.50\).
- Ratio of the powers is \(\frac{{{P_2}}}{{{P_1}}} = 4.0\).

**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.**

- 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: \({P_1} = {P_2}\).

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 \(4.0\).

(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 \(0.50\).

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 \(4.0\).

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