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Chapter 25: Problem 71

A \(12.0 \mathrm{~V}\) battery with an internal resistance \(R_{\mathrm{j}}=4.00 \Omega\) is attached across an external resistor of resistance \(R\). Find the maximum power that can be delivered to the resistor.

Short Answer

Expert verified
Answer: The maximum power that can be delivered to the external resistor is 9.00 W.

Step by step solution

01

Write down given values and formula for maximum power transfer theorem

Given values: Voltage of the battery, V = 12.0 V Internal resistance of the battery, \(R_j\) = 4.00 Ω Maximum power transfer theorem states that the maximum power is transferred to the load when the load resistance (R) is equal to the internal resistance of the source (\(R_j\)). Therefore, \(R = R_j\).
02

Calculate the equivalent resistance

Now, we will find the equivalent resistance of the circuit, which is the sum of the internal resistance and the external resistance (\(R_{eq} = R_j + R\)). Since \(R = R_j\), we can write: \(R_{eq} = R_j + R_j = 2R_j\) Plugging in the given value for \(R_j\): \(R_{eq} = 2(4.00 \Omega) = 8.00 \Omega\)
03

Calculate the current through the circuit

Now we can use Ohm's law to calculate the current (I) through the circuit: \(I = \frac{V}{R_{eq}}\) Substitute the values for V and \(R_{eq}\): \(I = \frac{12.0 V}{8.00 \Omega} = 1.50 \,\mathrm{A}\)
04

Calculate the maximum power delivered to the resistor

Finally, we can calculate the maximum power (P) that can be delivered to the resistor using the formula: \(P = I^2 R\) We know that \(R = R_j\), so: \(P = I^2 R_j\) Now, substitute the given values: \(P = (1.50 \,\mathrm{A})^2 (4.00 \Omega) = 2.25 \,\mathrm{A^2} (4.00 \Omega) = 9.00 \,\mathrm{W}\) The maximum power that can be delivered to the resistor is 9.00 W.

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Most popular questions from this chapter

One brand of \(12.0-\mathrm{V}\) automotive battery used to be advertised as providing " 600 cold-cranking amps." Assuming that this is the current the battery supplies if its terminals are shorted, that is, connected to negligible resistance, determine the internal resistance of the battery.

A charged-particle beam is used to inject a charge, \(Q_{0}\), into a small, irregularly shaped region (not a cavity, just some region within the solid block) in the interior of a block of ohmic material with conductivity \(\sigma\) and permittivity \(\epsilon\) at time \(t=0\). Eventually, all the injected charge will move to the outer surface of the block, but how quickly? a) Derive a differential equation for the charge, \(Q(t)\), in the injection region as a function of time. b) Solve the equation from part (a) to find \(Q(t)\) for all \(t \geq 0\). c) For copper, a good conductor, and for quartz (crystalline \(\mathrm{SiO}_{2}\) ), an insulator, calculate the time for the charge in the injection region to decrease by half. Look up the necessary values. Assume that the effective "dielectric constant" of copper is \(1.00000 .\)

A light bulb is connected to a source of emf. There is a \(6.20 \mathrm{~V}\) drop across the light bulb, and a current of 4.1 A flowing through the light bulb. a) What is the resistance of the light bulb? b) A second light bulb, identical to the first, is connected in series with the first bulb. The potential drop across the bulbs is now \(6.29 \mathrm{~V},\) and the current through the bulbs is \(2.9 \mathrm{~A}\). Calculate the resistance of each light bulb. c) Why are your answers to parts (a) and (b) not the same?

A rectangular wafer of pure silicon, with resistivity \(\rho=2300 \Omega \mathrm{m},\) measures \(2.00 \mathrm{~cm}\) by \(3.00 \mathrm{~cm}\) by \(0.010 \mathrm{~cm}\) Find the maximum resistance of this rectangular wafer between any two faces.

What is the resistance of a copper wire of length \(l=\) \(10.9 \mathrm{~m}\) and diameter \(d=1.3 \mathrm{~mm} ?\) The resistivity of copper is \(1.72 \cdot 10^{-8} \Omega \mathrm{m}\)
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