Question

A typical small flashlight contains two batteries, each having an emf of 1.5 V, connected in series with a bulb having resistance \(17 \Omega\). (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for 5.0 h, what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.)

Short answer

(a) 0.528 W; (b) 9504 J; (c) 6.0 Ω.

Step-by-step solution

Find Total EMF in Series

The flashlight has two batteries connected in series, each with an emf of 1.5 V. The total emf for batteries in series is the sum of the individual emf values: \[ E_{total} = E_1 + E_2 = 1.5\,V + 1.5\,V = 3.0\,V \]

Compute Current through the Bulb

Ohm's law states \( V = IR \), where \( V \) is the voltage across the bulb, \( I \) is the current through the bulb, and \( R \) is the resistance of the bulb. Because the internal resistance of the batteries is negligible, \( V = E_{total} \). Solving for current \( I \): \[ I = \frac{V}{R} = \frac{3.0\,V}{17\,\Omega} \approx 0.176\,A \]

Calculate the Power Delivered to Bulb

The power \( P \) delivered to a device is given by \( P = IV \). Using the current calculated and the total voltage: \[ P = 0.176\,A \times 3.0\,V = 0.528\,W \]

Determine Total Energy Delivered

Energy \( E \) delivered over a period \( t \) can be calculated with \( E = Pt \), where \( t \) is time in seconds. Given that time is 5 hours: \[ t = 5 \times 3600 = 18000\,s \] Thus, \[ E = 0.528\,W \times 18000\,s = 9504\,J \]

Find Combined Internal Resistance when Power Halves

When the power halves, the effective power is \( \frac{0.528}{2} = 0.264\,W \). Using the formula for power, \( P = I^2 R \), and solving for new current \( I' \), we have: \[ 0.264 = (I')^2 \times 17\,\Omega \Rightarrow I' = \sqrt{\frac{0.264}{17}} \approx 0.1257\,A \] Now, using \( V = E_{total} - I' r \) and \( P = I'^2 R \), \[ 3.0 = I'(17 + r) \] Solving, \[ 3.0 = 0.1257(17 + r) \Rightarrow r \approx 6.0\,\Omega. \]

Key concepts

Ohm's Law

Ohm’s Law is a fundamental principle used in the analysis of electric circuits. This law states that the current (I) through a conductor between two points is directly proportional to the voltage (V) across the two points. The mathematical representation is given by \[ V = I R \] where \( R \) is the resistance.
In our flashlight problem, this is essential for determining the current flowing through the bulb. By using the total voltage supplied by the series-connected batteries and the known resistance of the bulb, we can calculate the current. Ohm’s Law serves as a basic tool to relate voltage, current, and resistance, making it crucial for understanding various aspects of electrical circuits and problem-solving.

Power Calculations

Power in an electric circuit is the rate at which energy is delivered or consumed. It's calculated using the formula: \[ P = IV \] where \( P \) is power, \( I \) is current, and \( V \) is voltage.
In the flashlight example, this formula helps us calculate the power delivered to the bulb using the current from Ohm's Law and the total voltage. The steps show that with a current of approximately 0.176 A and a voltage of 3.0 V, the power is 0.528 W. This calculation is significant because it shows how efficiently the bulb is converting electrical energy into light and heat. Understanding power calculations allows us to evaluate the performance of electrical appliances.

Energy Delivered

Energy delivered in an electrical circuit is the total work done over a period of time. The formula to calculate energy is \[ E = Pt \] where \( E \) is energy in joules, \( P \) is power in watts, and \( t \) is time in seconds.
In the problem, the battery lasts for 5 hours. Converting this time into seconds gives 18000 seconds. The energy delivered to the bulb is then \( 0.528 \, W \times 18000 \, s = 9504 \, J \). This concept of energy calculation helps in understanding how long a device can run on its power source before needing a recharge or replacement. For practical applications, knowing how much energy an appliance uses is important for designing efficient systems and for cost calculations in power consumption.

Internal Resistance

Batteries have an internal resistance that affects the current they can supply. Initially, in the flashlight example, this resistance was negligible. However, as the batteries discharge, this resistance increases, affecting the power output.
To find the internal resistance when the power to the bulb halves, we use the power and Ohm’s Law equations. Given the initial power \( 0.528 \, W \) halving to \( 0.264 \, W \), and knowing that the bulb’s resistance is constant, we calculate the new current and use this to solve for the combined internal resistance, finding it to be approximately 6 \( \Omega \).
This concept helps to understand how battery efficiency decreases over time and why internal resistance is a critical factor in circuit design.

Series Circuit

A series circuit is one where components are connected end-to-end, so the current flows through each component sequentially. Total voltage is the sum of individual voltages, making it straightforward to calculate in simple configurations. In the flashlight problem, two batteries set in series provide a combined emf of 3.0 V (1.5 V each), directly affecting the bulb.
One key advantage of series circuits is simplicity, often used in applications where the same current must pass through each component, like string lights. However, if one part fails, the entire circuit is broken. Understanding how series circuits function and their uses is essential for designing effective electronic devices.