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Chapter 20: Problem 75

When a light bulb is connected across the terminals of a battery, the battery delivers 24 \(\mathrm{W}\) of power to the bulb. A voltage of 11.8 \(\mathrm{V}\) exists between the terminals of the battery, which has an internal resistance of 0.10\(\Omega\) . What is the emf of the battery?

Short Answer

Expert verified
The emf of the battery is approximately 12.0 V.

Step by step solution

01

Identify Given Values

We are given the power \( P = 24 \, \mathrm{W} \), the voltage across the battery terminals \( V = 11.8 \, \mathrm{V} \), and the internal resistance \( r = 0.10 \, \Omega \). We need to find the electromotive force (emf) \( \mathcal{E} \) of the battery.
02

Use Power Formula

The formula for power is given by \( P = I \cdot V \), where \( I \) is the current. Solve for the current \( I = \frac{P}{V} = \frac{24}{11.8} \approx 2.03 \, \mathrm{A} \).
03

Calculate Voltage Drop

The voltage drop across the internal resistance is found using Ohm's Law: \( V_r = I \cdot r = 2.03 \cdot 0.10 = 0.203 \, \mathrm{V} \).
04

Find Emf of the Battery

The emf of the battery is the sum of the terminal voltage and the voltage drop across the internal resistance: \( \mathcal{E} = V + V_r = 11.8 + 0.203 = 12.003 \, \mathrm{V} \).
05

Final Answer

The emf of the battery is approximately \( 12.0 \, \mathrm{V} \).

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Key Concepts

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

Internal Resistance
Internal resistance is an important concept in understanding how batteries function. It represents the inherent resistance to the flow of electric current within the battery itself. When a battery is in use, the electrons encounter this resistance while passing through the battery.

Key points to remember about internal resistance:
  • It causes a voltage drop inside the battery, meaning not all the battery's energy is available to the external circuit.
  • Internal resistance, denoted by the symbol \( r \), is measured in ohms (惟).
  • This resistance affects the efficiency of a battery. The higher the internal resistance, the less efficient the battery is.
Imagine it like a hurdle inside the battery 鈥� the more hurdles, the tougher it is for the current to flow freely. In this exercise, the internal resistance was given at 0.10 惟, contributing to a slight voltage drop before the energy reaches the connected light bulb.
Ohm's Law
Ohm's Law is the backbone of many electrical calculations. It's a simple but crucial relationship between voltage, current, and resistance. The law is typically expressed as \( V = I \cdot R \), where:
  • \( V \) is the voltage across the conductor in volts (V),
  • \( I \) is the current through the conductor in amperes (A), and
  • \( R \) is the resistance of the conductor in ohms (惟).
In the context of this exercise, Ohm's Law helps us calculate the voltage drop across the internal resistance of the battery. By using the formula \( V_r = I \cdot r \), we determined the voltage drop was 0.203 V. This drop is critical as it shows how internal resistance impacts the effective voltage, highlighting the losses before energy reaches the external circuit.
Electric Power
Electric power is the rate at which electrical energy is transferred by an electric circuit. The unit of power is the watt (W). The formula to calculate power is \( P = I \cdot V \), where \( P \) is power, \( I \) is current, and \( V \) is voltage.

Here's why understanding electric power is vital:
  • It provides insights into how much energy is being used by a device, like a light bulb in our case.
  • Power indicates the rate of energy consumption or production, which is crucial for calculating the cost of electricity.
In this exercise, the light bulb needed 24 W of power, reflecting how swiftly it uses the energy supplied by the battery. By rearranging the power equation, we solved for the current and found it to be approximately 2.03 A, essential for further calculations.
Battery Voltage
Battery voltage is a critical measure of a battery鈥檚 ability to provide energy. It's the potential difference between the battery's terminals when no current flows (known as electromotive force or emf) and varies due to several factors.

Key aspects of battery voltage include:
  • The emf, denoted by \( \mathcal{E} \), is the maximum potential difference a battery can deliver when no current is drawn.
  • The terminal voltage is what you measure across the terminals while the battery is powering a device and may be less than the emf due to internal resistance.
  • Understanding both emf and terminal voltage helps in assessing battery efficiency and functionality.
In this problem, the battery had a terminal voltage of 11.8 V, and internal resistance caused a small decrease from the emf. After factoring in the drop due to internal resistance, the battery's emf was computed to be approximately 12.0 V, signifying its full potential energy capability.

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Most popular questions from this chapter

The coil of wire in a galvanometer has a resistance of \(R_{C}=60.0 \Omega\) . The galvanometer exhibits a full-scale deflection when the current through it is 0.400 \(\mathrm{mA}\) . A resistor is connected in series with this combination so as to produce a nondigital voltmeter. The voltmeter is to have a full-scale deflection when it measures a potential difference of 10.0 \(\mathrm{V}\) . What is the resistance of this resistor?

A coffee-maker \((14 \Omega)\) and a toaster \((19 \Omega)\) are connected in parallel to the same \(120-\mathrm{V}\) outlet in a kitchen. How much total power is supplied to the two appliances when both are turned on?

The total current delivered to a number of devices connected in parallel is the sum of the individual currents in each device. Circuit breakers are resettable automatic switches that protect against a dangerously large total current by 鈥渙pening鈥� to stop the current at a specified safe value. A 1650-W toaster, a 1090-W iron, and a 1250-W microwave oven are turned on in a kitchen. As the drawing shows, they are all connected through a 20-A circuit breaker (which has negligible resistance) to an ac voltage of 120 V. (a) Find the equivalent resistance of the three devices. (b) Obtain the total current delivered by the source and determine whether the breaker will 鈥渙pen鈥� to prevent an accident.

The coil of a galvanometer has a resistance of \(20.0 \Omega,\) and its meter deflects full scale when a current of 6.20 \(\mathrm{mA}\) passes through it. To make the galvanometer into a nondigital ammeter, a 24.8 \(\mathrm{m} \Omega\) shunt resistor is added to it. What is the maximum current that this ammeter can read?

Two capacitors are connected to a battery. The battery voltage is \(V=60.0 \mathrm{V}\) , and the capacitances are \(C_{1}=2.00 \mu \mathrm{F}\) and \(C_{2}=4.00 \mu \mathrm{F}\) . Determine the total energy stored by the two capacitors when they are wired \((\mathrm{a})\) in parallel and \((\mathrm{b})\) in series.
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