10.1 Electromotive force  (Page 6/11)

 Page 6 / 11

Summary

• All voltage sources have two fundamental parts: a source of electrical energy that has a characteristic electromotive force (emf), and an internal resistance r . The emf is the work done per charge to keep the potential difference of a source constant. The emf is equal to the potential difference across the terminals when no current is flowing. The internal resistance r of a voltage source affects the output voltage when a current flows.
• The voltage output of a device is called its terminal voltage ${V}_{\text{terminal}}$ and is given by ${V}_{\text{terminal}}=\epsilon -Ir$ , where I is the electric current and is positive when flowing away from the positive terminal of the voltage source and r is the internal resistance.

Conceptual questions

What effect will the internal resistance of a rechargeable battery have on the energy being used to recharge the battery?

Some of the energy being used to recharge the battery will be dissipated as heat by the internal resistance.

A battery with an internal resistance of r and an emf of 10.00 V is connected to a load resistor $R=r$ . As the battery ages, the internal resistance triples. How much is the current through the load resistor reduced?

Show that the power dissipated by the load resistor is maximum when the resistance of the load resistor is equal to the internal resistance of the battery.

$\begin{array}{}\\ \\ P={I}^{2}R={\left(\frac{\epsilon }{r+R}\right)}^{2}R={\epsilon }^{2}R{\left(r+R\right)}^{-2},\phantom{\rule{0.5em}{0ex}}\frac{dP}{dR}={\epsilon }^{2}\left[{\left(r+R\right)}^{-2}-2R{\left(r+R\right)}^{-3}\right]=0,\hfill \\ \left[\frac{\left(r+R\right)-2R}{{\left(r+R\right)}^{3}}\right]=0,\phantom{\rule{0.5em}{0ex}}r=R\hfill \end{array}$

Problems

A car battery with a 12-V emf and an internal resistance of $0.050\phantom{\rule{0.2em}{0ex}}\text{Ω}$ is being charged with a current of 60 A. Note that in this process, the battery is being charged. (a) What is the potential difference across its terminals? (b) At what rate is thermal energy being dissipated in the battery? (c) At what rate is electric energy being converted into chemical energy?

The label on a battery-powered radio recommends the use of rechargeable nickel-cadmium cells (nicads), although they have a 1.25-V emf, whereas alkaline cells have a 1.58-V emf. The radio has a $3.20\phantom{\rule{0.2em}{0ex}}\text{Ω}$ resistance. (a) Draw a circuit diagram of the radio and its batteries. Now, calculate the power delivered to the radio (b) when using nicad cells, each having an internal resistance of $0.0400\phantom{\rule{0.2em}{0ex}}\text{Ω}$ , and (c) when using alkaline cells, each having an internal resistance of $0.200\phantom{\rule{0.2em}{0ex}}\text{Ω}$ . (d) Does this difference seem significant, considering that the radio’s effective resistance is lowered when its volume is turned up?

a.

b. 0.476W; c. 0.691 W; d. As ${R}_{L}$ is lowered, the power difference decreases; therefore, at higher volumes, there is no significant difference.

An automobile starter motor has an equivalent resistance of $0.0500\phantom{\rule{0.2em}{0ex}}\text{Ω}$ and is supplied by a 12.0-V battery with a $0.0100\text{-}\text{Ω}$ internal resistance. (a) What is the current to the motor? (b) What voltage is applied to it? (c) What power is supplied to the motor? (d) Repeat these calculations for when the battery connections are corroded and add $0.0900\phantom{\rule{0.2em}{0ex}}\text{Ω}$ to the circuit. (Significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.)

(a) What is the internal resistance of a voltage source if its terminal potential drops by 2.00 V when the current supplied increases by 5.00 A? (b) Can the emf of the voltage source be found with the information supplied?

a. $0.400\phantom{\rule{0.2em}{0ex}}\text{Ω}$ ; b. No, there is only one independent equation, so only r can be found.

A person with body resistance between his hands of $10.0\phantom{\rule{0.2em}{0ex}}\text{k}\text{Ω}$ accidentally grasps the terminals of a 20.0-kV power supply. (Do NOT do this!) (a) Draw a circuit diagram to represent the situation. (b) If the internal resistance of the power supply is $2000\phantom{\rule{0.2em}{0ex}}\text{Ω}$ , what is the current through his body? (c) What is the power dissipated in his body? (d) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in this situation to be 1.00 mA or less? (e) Will this modification compromise the effectiveness of the power supply for driving low-resistance devices? Explain your reasoning.

A 12.0-V emf automobile battery has a terminal voltage of 16.0 V when being charged by a current of 10.0 A. (a) What is the battery’s internal resistance? (b) What power is dissipated inside the battery? (c) At what rate (in $\text{°}\text{C}\text{/}\text{min}$ ) will its temperature increase if its mass is 20.0 kg and it has a specific heat of $0.300\phantom{\rule{0.2em}{0ex}}\text{kcal/kg}\phantom{\rule{0.2em}{0ex}}·\text{°}\text{C}$ , assuming no heat escapes?

a. $0.400\phantom{\rule{0.2em}{0ex}}\text{Ω}$ ; b. 40.0 W; c. $0.0956\phantom{\rule{0.2em}{0ex}}\text{°C/min}$

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