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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

The resistivity of a conductor is \(\rho=1.00 \cdot 10^{-5} \Omega \mathrm{m}\). If a cylindrical wire is made of this conductor, with a crosssectional area of \(1.00 \cdot 10^{-6} \mathrm{~m}^{2},\) what should the length of the wire be for its resistance to be \(10.0 \Omega ?\)

In an emergency, you need to run a radio that uses \(30.0 \mathrm{~W}\) of power when attached to a \(10.0-\mathrm{V}\) power supply. The only power supply you have access to provides \(25.0 \mathrm{kV}\) but you do have a large number of \(25.0-\Omega\) resistors. If you want the power to the radio to be as close as possible to \(30.0 \mathrm{~W}\), how many resistors should you use, and how should they be connected (in series or in parallel)?

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A battery has a potential difference of \(14.50 \mathrm{~V}\) when it is not connected in a circuit. When a \(17.91-\Omega\) resistor is connected across the battery, the potential difference of the battery drops to \(12.68 \mathrm{~V}\). What is the internal resistance of the battery?

A constant electric field is maintained inside a semiconductor. As the temperature is lowered, the magnitude of the current density inside the semiconductor a) increases. c) decreases. b) stays the same. d) may increase or decrease.
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