Problem 45 - University Physics with Modern Physics

Question

A 25.0-\(\Omega\) bulb is connected across the terminals of a 12.0-V battery having 3.50\(\Omega\) of internal resistance. What percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb?

Step-by-Step Solution

Answer

12.28% of the power is dissipated across the internal resistance.

Step 1: Calculate Total Resistance

First, find the total circuit resistance by adding the bulb's resistance to the internal resistance of the battery. \[ R_{total} = R_{bulb} + R_{internal} = 25.0\,\Omega + 3.50\,\Omega = 28.5\,\Omega \]

Step 2: Calculate Total Current

Use Ohm's law to calculate the total current flowing through the circuit. \[ I = \frac{V}{R_{total}} = \frac{12.0\,V}{28.5\,\Omega} \approx 0.421\,A \]

Step 3: Calculate Power Dissipated in Internal Resistance

Use the formula for power \( P = I^2 R \) to find the power dissipated across the internal resistance. \[ P_{internal} = (0.421\,A)^2 \times 3.50\,\Omega \approx 0.620\,W \]

Step 4: Calculate Total Power Provided by the Battery

Calculate the total power provided by the battery using the formula \( P = IV \). \[ P_{total} = 0.421\,A \times 12.0\,V \approx 5.05\,W \]

Step 5: Calculate Percentage of Power Dissipated in Internal Resistance

Calculate the percentage of total power that is dissipated in the internal resistance using the formula \( \text{Percentage} = \left( \frac{P_{internal}}{P_{total}} \right) \times 100 \% \).\[ \text{Percentage} = \left( \frac{0.620\,W}{5.05\,W} \right) \times 100 \% \approx 12.28\% \]

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle used to analyze electrical circuits. It defines the relationship between voltage, current, and resistance. According to Ohm's Law, the current flowing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance of the conductor. This can be expressed by the formula:

• \( V = IR \) where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance.

In our exercise, Ohm's Law is used to calculate the current flowing through the circuit with a battery voltage of 12.0 V and a total resistance of 28.5 \( \Omega \). Applying Ohm’s Law:

• \( I = \frac{V}{R} = \frac{12.0\,V}{28.5\,\Omega} \approx 0.421\,A \)

This means approximately 0.421 amperes of current flows through the circuit. Understanding Ohm's Law helps in predicting how changes in voltage or resistance can affect current in a circuit.

Power Dissipation

Power dissipation refers to the process of electrical energy being converted into thermal energy in a circuit. This energy is usually lost as heat, affecting the efficiency of the circuit.To find power dissipation, the formula \( P = I^2R \) is used, where:

• \( P \) is the power in watts,
• \( I \) is the current in amperes, and
• \( R \) is the resistance in ohms.

In the context of this exercise, we calculate the power dissipated by the internal resistance of the battery. Using the previously calculated current of 0.421 A, the power dissipated is:

• \( P_{internal} = (0.421\,A)^2 \times 3.50\,\Omega \approx 0.620\,W \)

This tells us that approximately 0.620 watts of power is lost due to the internal resistance, and it is not available to power the bulb. Calculating power dissipation is vital for efficient circuit design.

Internal Resistance

Internal resistance is the inherent resistance to the flow of current within a battery or power source. This resistance causes a portion of the energy provided by the battery to be lost as heat rather than fully delivering the voltage to the external circuit.In a real battery, internal resistance means that the terminal voltage is slightly less than the actual voltage produced inside the battery. The formula for internal resistance's contribution to total voltage can be understood as:

• \( V_{terminal} = V - IR_{internal} \)

where \( V \) is the voltage of the battery, \( I \) is the current, and \( R_{internal} \) is the internal resistance.In our example, the internal resistance of 3.50 \( \Omega \) contributed to the power being dissipated as heat, calculated as 0.620 W. Consequently, this reduced the power available to the bulb itself. By understanding internal resistance, one can better evaluate and design circuits to minimize energy loss, ensure optimal performance, and maintain battery life.