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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Battery Internal Resistance
Understanding the concept of internal resistance is crucial when dealing with batteries in electric circuits. Batteries are not perfect power sources; they have an intrinsic resistance termed as 'internal resistance'. This resistance is due mainly to the materials and construction of the battery which impedes the flow of electric current.

Here is what you need to know about internal resistance:
  • Internal resistance ( ) is measured in ohms (Ω) and it influences the current ( I) that a battery can supply.
  • When a battery supplies current, some of its voltage is spent overcoming this internal resistance, decreasing the effective voltage in the circuit.
This means that the voltage across a battery's terminals is always less than its electromotive force (emf), described by the equation: \[V = V_{emf} - Ir\]Here, V is the voltage across the battery, V_{emf} is the battery’s emf, I is the current, and r is the internal resistance. Recognizing this distinction helps prevent misunderstandings in circuit calculations.
Power in Circuits
Power in electrical circuits refers to the rate at which electrical energy is converted to another form, such as heat, light, or mechanical energy. In the context of circuits, power can be determined through several essential formulas, one of which is \[P = \frac{V^2}{R}\]where \(P\) represents the power in watts, \(V\) is the voltage across the component, and \(R\) is the resistance.

Key considerations when calculating power in circuits include:
  • The power formula is directly associated with Ohm's law, implying close attention must be paid to the relationship between voltage, current, and resistance.
  • Extracting power from a circuit involves converting the voltage supplied by the battery, taking into account the total resistance affecting the current flow.
In practical applications, knowing how to calculate the power allows you to design circuits efficiently by selecting appropriate components to manage power loads without overheating.
Ohm's Law
Ohm's Law is a fundamental principle in the study of electric circuits, connecting voltage, current, and resistance in a simple relation. It is commonly expressed as: \[V = IR\]in which \(V\) stands for voltage, \(I\) indicates current, and \(R\) denotes resistance.

This law forms the backbone of circuit analysis, helping us understand how electricity flows through a conductor. Here are some points to note:
  • When dealing with circuits, changes in resistance and voltage directly affect the current passing through the circuit, according to Ohm's Law.
  • Ohm's Law helps verify circuit behavior, predicting how variables change as conditions in the circuit are altered.
Understanding and applying Ohm's Law is essential for solving problems involving multiple components in a circuit. It allows you to discern the role each plays in the entire system, making it indispensable in circuit design and troubleshooting.

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Most popular questions from this chapter

Which of the following will reduce the time constant in an RC circuit? a) increasing the dielectric constant of the capacitor b) adding an additional \(20 \mathrm{~m}\) of wire between the capacitor and the resistor c) increasing the voltage of the battery d) adding an additional resistor in parallel with the first resistor e) none of the above

The dead battery of your car provides a potential difference of \(9.950 \mathrm{~V}\) and has an internal resistance of \(1.100 \Omega\). You charge it by connecting it with jumper cables to the live battery of another car. The live battery provides a potential difference of \(12.00 \mathrm{~V}\) and has an internal resistance of \(0.0100 \Omega\), and the starter resistance is \(0.0700 \Omega\). a) Draw the circuit diagram for the connected batteries. b) Determine the current in the live battery, in the dead battery, and in the starter immediately after you closed the circuit.

Voltmeters are always connected in parallel with a circuit component, and ammeters are always connected in series. Explain why.

An ammeter with an internal resistance of \(53 \Omega\) measures a current of \(5.25 \mathrm{~mA}\) in a circuit containing a battery and a total resistance of \(1130 \Omega\). The insertion of the ammeter alters the resistance of the circuit, and thus the measurement does not give the actual value of the current in the circuit without the ammeter. Determine the actual value of the current.

How long would it take, in multiples of the time constant, \(\tau,\) for the capacitor in an \(\mathrm{RC}\) circuit to be \(98 \%\) charged? a) \(9 \tau\) c) \(90 \tau\) e) \(0.98 \tau\) b) \(0.9 \tau\) d) \(4 \tau\)

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