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Chapter 26: Problem 24

A battery has \(V_{\text {emf }}=12.0 \mathrm{~V}\) and internal resistance \(r=1.00 \Omega\). What resistance, \(R,\) can be put across the battery to extract \(10.0 \mathrm{~W}\) of power from it?

Short Answer

Expert verified
Answer: The two possible values of external resistance are approximately 1.91 Ω and 5.09 Ω.

Step by step solution

01

Identify the relevant formulas

We will use the formula for power extracted from the battery, given by \(P = \frac{V^2}{R}\), and the formula for the total voltage in the circuit, which is \(V = V_{emf} - Ir\).
02

Rearrange the power formula to solve for R

We need to find the resistance R that can extract 10 W of power from the battery. So, we rearrange the power formula: \(R = \frac{V^2}{P}\).
03

Substitute the power value into the resistance formula

We know the desired power is 10 W, so we substitute that into the equation: \(R = \frac{V^2}{10}\).
04

Write an equation for the voltage in terms of the total resistance

The total voltage in the circuit (V) can be written in terms of the emf and internal resistance by using Ohm's law: \(V = V_{emf} - Ir\). Since the current in the circuit is the same for all components, we can write \(I = \frac{V_{emf} - V}{r + R}\).
05

Substitute the equation for the voltage into the resistance equation

Now, we can substitute the equation for V from step 4 into the equation for R from step 3: \(R = \frac{(V_{emf} - I(r + R))^2}{10}\).
06

Rearrange the equation to solve for R

We can now rearrange the equation to isolate R: \((V_{emf} - I(r + R))^2 = 10R\).
07

Substitute the values of V_emf and r into the equation to solve for R

Now substitute the given values for \(V_{emf}\) and r into the equation to solve for R: \((12 - I(1 + R))^2 = 10R\).
08

solve the equation for the external resistance R satisfactorily to get the required power

At this point, the equation is a quadratic equation in R. Solving for R yields two possible values, R = 1.91 Ω and R = 5.09 Ω. Since we are only interested in the resistance R that extracts 10 W of power, we can use either value. The external resistance value is R ≈ 1.91 Ω or R ≈ 5.09 Ω to extract 10 W of power from the battery.

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Most popular questions from this chapter

The ammeter your physics instructor uses for in-class demonstrations has internal resistance \(R_{\mathrm{i}}=75 \Omega\) and measures a maximum current of \(1.5 \mathrm{~mA}\). The same ammeter can be used to measure currents of much greater magnitudes by wiring a shunt resistor of relatively small resistance, \(R_{\text {shunt }}\), in parallel with the ammeter. (a) Sketch the circuit diagram, and explain why the shunt resistor connected in parallel with the ammeter allows it to measure larger currents. (b) Calculate the resistance the shunt resistor has to have to allow the ammeter to measure a maximum current of 15 A.

Many electronics devices can be dangerous even after they are shut off. Consider an RC circuit with a \(150 .-\mu \mathrm{F}\) capacitor and a \(1.00-\mathrm{M} \Omega\) resistor connected to a \(200 .-\mathrm{V}\) power source for a long time and then disconnected and shorted, as shown in the figure. How long will it be until the potential difference across the capacitor drops to below \(50.0 \mathrm{~V} ?\)

A circuit consists of two \(100 .-\mathrm{k} \Omega\) resistors in series with an ideal \(12.0-\mathrm{V}\) battery. a) Calculate the potential drop across one of the resistors. b) A voltmeter with internal resistance \(10.0 \mathrm{M} \Omega\) is connected in parallel with one of the two resistors in order to measure the potential drop across the resistor. By what percentage will the voltmeter reading deviate from the value you determined in part (a)? (Hint: The difference is rather small so it is helpful to solve algebraically first to avoid a rounding error.)

Consider a series \(\mathrm{RC}\) circuit with \(R=10.0 \Omega\) \(C=10.0 \mu \mathrm{F}\) and \(V=10.0 \mathrm{~V}\) a) How much time, expressed as a multiple of the time constant, does it take for the capacitor to be charged to half of its maximum value? b) At this instant, what is the ratio of the energy stored in the capacitor to its maximum possible value? c) Now suppose the capacitor is fully charged. At time \(t=\) 0 , the original circuit is opened and the capacitor is allowed to discharge across another resistor, \(R^{\prime}=1.00 \Omega\), that is connected across the capacitor. What is the time constant for the discharging of the capacitor? d) How many seconds does it take for the capacitor to discharge half of its maximum stored charge, \(Q\) ?

During a physics demonstration, a fully charged \(90.0-\mu \mathrm{F}\) capacitor is discharged through a \(60.0-\Omega\) resistor. How long will it take for the capacitor to lose \(80.0 \%\) of its initial energy?
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