Question

Three resistors having resistances of 1.60 \(\Omega\), 2.40 \(\Omega\), and 4.80 \(\Omega\) are connected in parallel to a 28.0-V battery 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; (e) the power dissipated in each resistor. (f) Which resistor dissipates the most power: the one with the greatest resistance or the least resistance? Explain why this should be.

Short answer

(a) 0.80 Ω, (b) 17.5 A, 11.67 A, 5.83 A, (c) 35 A, (d) 28.0 V, (e) 490 W, 327 W, 163 W, (f) The least resistance dissipates the most power.

Step-by-step solution

Find the Equivalent Resistance

When resistors are in parallel, the equivalent resistance \( R_{eq} \) is given by the formula: \[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \] where \( R_1 = 1.60 \, \Omega \), \( R_2 = 2.40 \, \Omega \), and \( R_3 = 4.80 \, \Omega \). Substituting the given values:\[ \frac{1}{R_{eq}} = \frac{1}{1.60} + \frac{1}{2.40} + \frac{1}{4.80} \]Calculating the reciprocals:\[ \frac{1}{R_{eq}} = 0.625 + 0.417 + 0.208 \]\[ \frac{1}{R_{eq}} = 1.25 \]Finally, the equivalent resistance is:\[ R_{eq} = \frac{1}{1.25} = 0.80 \Omega \].

Find the Current through Each Resistor

The current through a resistor in a parallel circuit can be calculated using Ohm's Law \( I = \frac{V}{R} \) where \( V = 28.0 \, V \) is the voltage across each resistor.For \( R_1 = 1.60 \, \Omega \): \[ I_1 = \frac{28.0}{1.60} = 17.5 \, A \]For \( R_2 = 2.40 \, \Omega \): \[ I_2 = \frac{28.0}{2.40} = 11.67 \, A \]For \( R_3 = 4.80 \, \Omega \):\[ I_3 = \frac{28.0}{4.80} = 5.83 \, A \].

Find the Total Current through the Battery

The total current through the battery is the sum of the currents through each resistor:\[ I_{total} = I_1 + I_2 + I_3 \]\[ I_{total} = 17.5 + 11.67 + 5.83 = 35 \, A \].

Check the Voltage across Each Resistor

In a parallel circuit, the voltage across each resistor is equal to the voltage of the battery. Therefore, the voltage across each of the resistors is \( 28.0 \, V \).

Calculate the Power Dissipated in Each Resistor

The power dissipated in a resistor can be calculated using the formula \( P = I^2 R \).For \( R_1 \):\[ P_1 = (17.5)^2 \times 1.60 = 490 \, W \]For \( R_2 \):\[ P_2 = (11.67)^2 \times 2.40 = 327 \, W \]For \( R_3 \):\[ P_3 = (5.83)^2 \times 4.80 = 163 \, W \].

Determine which Resistor Dissipates the Most Power

The resistor with the least resistance (\( R_1 = 1.60 \, \Omega \)) dissipates the most power since it has the highest current flow, following the relationship \( P = I^2 R \). Lower resistance in parallel circuits leads to higher current, causing greater power dissipation despite the higher resistance values.

Key concepts

Ohm's Law

Ohm's Law is a basic principle in electronics that helps us understand the relationship between voltage, current, and resistance. It's expressed with the formula \( V = IR \).
In this formula:

• \( V \) represents voltage (in volts). It is the potential difference that makes electric charges move.
• \( I \) denotes current (in amperes). It measures the flow of electric charge.
• \( R \) stands for resistance (in ohms). It indicates how much a material resists the flow of current.

In a parallel circuit, like in this exercise, the voltage across each component is equal. Therefore, when calculating the current through each resistor using Ohm’s Law, we use the formula \( I = \frac{V}{R} \).
This shows how current varies with different resistances while keeping voltage constant. In simpler terms, if you know two of these values, you can find the third!

Equivalent Resistance

In circuits, equivalent resistance is the single resistance that can replace a combination of resistors, giving the same effect on current flow. For resistors in parallel, calculating equivalent resistance is vital to simplifying the circuit analysis. The formula used here is:\[\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\]Where \( R_1, R_2, R_3 \) are the resistances of different components. By calculating reciprocals of the resistances, we find that:\[R_{eq} = \frac{1}{1.25} = 0.80\, \Omega\]This value, \( 0.80 \, \Omega \), is smaller than the smallest individual resistor in the parallel setup. Why is that important? In a parallel circuit, having multiple paths for current leads to lower overall resistance, allowing more current to flow.

Electrical Power

Electrical power quantifies the rate at which energy is used or transferred in a circuit. It's calculated as the product of current and voltage but can also be found using resistance and current:\[ P = I^2R \]In this exercise:

• The power dissipated in the 1.60 \( \Omega \) resistor is \( 490 \, W \), as it carries the highest current.
• The other resistors dissipate \( 327 \, W \) and \( 163 \, W \) respectively.

These calculations help us see how resistors convert electrical energy into heat. Notably, even with different resistances, the resistor with the lowest resistance dissipates the most power because of higher current (from Ohm’s Law \( I = \frac{V}{R} \)).
In simple terms, less resistance in a parallel circuit allows more current, leading to more energy (power) converted to heat.

Resistor Current Calculation

Determining the current through each resistor in a parallel circuit is straightforward using Ohm’s Law without needing complex math:
1. Calculate each current using \( I = \frac{V}{R} \) with \( V = 28.0 \, V \). For each resistor in this exercise:

• \( I_1 = \frac{28.0}{1.60} = 17.5 \, A \)
• \( I_2 = \frac{28.0}{2.40} = 11.67 \, A \)
• \( I_3 = \frac{28.0}{4.80} = 5.83 \, A \)

2. Total current through the battery is the sum of these currents: \[ I_{total} = I_1 + I_2 + I_3 = 35 \, A \]Understanding these steps shows that even in real-world applications, current divides according to the resistance across each path in a parallel circuit. This division allows for varying current amounts, depending on each pathway's resistance.