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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

How long will it take for the current in a circuit to drop from its initial value to \(1.50 \mathrm{~mA}\) if the circuit contains two \(3.8-\mu \mathrm{F}\) capacitors that are initially uncharged, two \(2.2-\mathrm{k} \Omega\) resistors, and a \(12.0-\mathrm{V}\) battery all connected in series?

A capacitor bank is designed to discharge 5.0 J of energy through a \(10.0-\mathrm{k} \Omega\) resistor array in under \(2.0 \mathrm{~ms}\) To what potential difference must the bank be charged, and what must the capacitance of the bank be?

A circuit consists of a source of emf, a resistor, and a capacitor, all connected in series. The capacitor is fully charged. How much current is flowing through it? a) \(i=V / R\) b) zero c) neither (a) nor (b)

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.)

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