Problem 50 - College Physics

Question

Three resistors having resistances of \(1.60 \Omega, 2.40 \Omega,\) and \(4.80 \Omega,\) respectively, are connected in parallel to a 28.0 \(\mathrm{V}\) bat- tery that has negligible internal resistance. Find (a) the equivalent resistance of the combination, (b) the current in each resistor, (c) the total current through the battery, (d) the voltage across each resistor, and (e) the power dissipated in each resistor. (f) Which resistor dissipates the most power, the one with the greatest resistance or the one with the least resistance? Explain why this should be.

Step-by-Step Solution

Answer

The equivalent resistance is 0.96 Ω. The currents are 17.5 A, 11.67 A, and 5.83 A. Total current is 35.0 A. Each voltage is 28.0 V. Powers are 490 W, 326.67 W, and 163.33 W. The lowest resistance ( 1.60 Ω) dissipates the most power.

Step 1: Understand the Problem

Read the problem carefully and identify that we need to find the equivalent resistance, the current across each resistor, the total current, the voltage across each resistor, the power dissipated by each resistor, and compare the powers dissipated.

Step 2: Calculate Equivalent Resistance

For resistors in parallel, the equivalent resistance \( R_{eq} \) is given by: \[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \] Plug in the values: \[ \frac{1}{R_{eq}} = \frac{1}{1.60} + \frac{1}{2.40} + \frac{1}{4.80} \] Simplifying gives \( R_{eq} = 0.96 \Omega \).

Step 3: Calculate Current in Each Resistor

Using Ohm's Law \( I = \frac{V}{R} \), where \( V = 28.0 \text{ V} \), compute the current through each resistor: - For \( R_1 = 1.60 \Omega: \) \[ I_1 = \frac{28.0}{1.60} = 17.5 \text{ A} \]- For \( R_2 = 2.40 \Omega: \) \[ I_2 = \frac{28.0}{2.40} = 11.67 \text{ A} \]- For \( R_3 = 4.80 \Omega: \) \[ I_3 = \frac{28.0}{4.80} = 5.83 \text{ A} \]

Step 4: Calculate Total Current Through Battery

The total current through the battery is the sum of the currents through each resistor: \[ I_{total} = I_1 + I_2 + I_3 = 17.5 + 11.67 + 5.83 = 35.0 \text{ A} \]

Step 5: Verify Voltage Across Each Resistor

Since the resistors are in parallel, the voltage across each resistor is the same as the battery voltage, which is \( 28.0 \text{ V} \).

Step 6: Calculate Power Dissipated in Each Resistor

Using the formula \( P = IV \) or \( P = \frac{V^2}{R} \), compute the power for each:- For \( R_1: \) \[ P_1 = \frac{28.0^2}{1.60} = 490 \text{ W} \]- For \( R_2: \) \[ P_2 = \frac{28.0^2}{2.40} = 326.67 \text{ W} \]- For \( R_3: \) \[ P_3 = \frac{28.0^2}{4.80} = 163.33 \text{ W} \]

Step 7: Analyze Most Power Dissipated

The resistor with the least resistance (\( R_1 = 1.60 \Omega \)) dissipates the most power. This is because, with a constant voltage, power dissipation \( P \) is inversely proportional to resistance \( R \), leading to higher power dissipation in lower resistance.

Key Concepts

Resistors in Parallel

In an electric circuit, resistors can be arranged in different ways, with parallel being one of the most common configurations. When resistors are connected in parallel, each one is hooked up across the same two points in the circuit. This means that each resistor experiences the same voltage as the total supply voltage, such as the one provided by a battery.

To find the equivalent resistance of resistors in parallel, which is important for analyzing the circuit, use the formula:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \, ... \, + \frac{1}{R_n} \]

Thus, the more resistors you add in parallel, the lower the equivalent resistance. This is because the circuit has more paths for the electric current to flow through, making it easier for the entire current to pass through. Always remember that the equivalent resistance of a parallel circuit is always less than the smallest resistor in that grouping.

Ohm's Law

Ohm's Law is fundamental to understanding how electric circuits work. This law states that the current passing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance.

The formula is quite simple:
\( I = \frac{V}{R} \)

Where:

• \( I \) is the current in amperes (A)
• \( V \) is the voltage in volts (V)
• \( R \) is the resistance in ohms (Ω)

By applying Ohm's Law, you can easily calculate the current flowing through each resistor in a parallel circuit by using the individual resistance values and the known voltage. This shows us how changes in resistance or voltage can affect the current flowing in a circuit.

Power Dissipation

Power dissipation refers to the conversion of electrical energy into heat energy in a resistor as current flows through it. Calculating power can tell us how much energy is used in a circuit over time.

The power dissipated by a resistor can be found using multiple formulas, dependent on what quantities you know:

• \( P = IV \)
• \( P = I^2R \)
  
• \( P = \frac{V^2}{R} \)

Each formula highlights different aspects of how power relates to other circuit attributes such as voltage, current, and resistance. In the case of resistors in parallel, knowing either the voltage or the current can aid in determining how much power is lost as heat in each resistor.

Equivalent Resistance

Equivalent resistance in a parallel circuit helps us understand how easily the total circuit accommodates current even though it has several resistors. Equivalent resistance describes the total resistance faced by the current as if all the parallel resistors were replaced by a single resistor.

This concept is useful for simplifying circuit analysis, allowing us to replace a complex section with a simpler representation. As more resistors are added in parallel, the equivalent resistance decreases.

In practical terms, this means if a circuit designer wants to reduce the overall resistance of a section of a circuit drastically, they can add more resistors in parallel. Lower equivalent resistance means that the circuit can carry more current overall, which is often beneficial in practical applications such as distributing power efficiently.