Question

The electronics supply company where you work has two different resistors, \(R_1\) and \(R_2\), in its inventory, and you must measure the values of their resistances. Unfortunately, stock is low, and all you have are \(R_1\) and \(R_2\) in parallel and in series-and you can't separate these two resistor combinations. You separately connect each resistor network to a battery with emf 48.0 V and negligible internal resistance and measure the power \(P\) supplied by the battery in both cases. For the series combination, \(P =\) 48.0 W; for the parallel combination, \(P =\) 256 W. You are told that \(R_1\) \(R_2\). (a) Calculate \(R_1\) and \(R_2\). (b) For the series combination, which resistor consumes more power, or do they consume the same power? Explain. (c) For the parallel combination, which resistor consumes more power, or do they consume the same power?

Short answer

(a) \(R_1 = 36\; \Omega\), \(R_2 = 12\; \Omega\); (b) \(R_1\) consumes more power in series; (c) \(R_2\) consumes more power in parallel.

Step-by-step solution

Interpret the Power in Series

For the series combination, the total resistance is the sum of individual resistances: \(R_s = R_1 + R_2\). Ohm's Law gives us the total resistance in terms of power and voltage: \(R_s = \frac{V^2}{P} = \frac{48^2}{48} = 48\; \Omega\).

Interpret the Power in Parallel

For the parallel combination, the total resistance \(R_p\) can be found by \(R_p = \frac{V^2}{P} = \frac{48^2}{256} = 9\; \Omega\).

Apply the Parallel Resistance Formula

In a parallel circuit, \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \). Setting this equation gives us: \(\frac{1}{9} = \frac{1}{R_1} + \frac{1}{R_2}\).

Set Up a System of Equations

The equations are: (1) \(R_1 + R_2 = 48\) and (2) \(\frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{9}\).

Solve the System of Equations

From equation (1), express \(R_2\) as \(R_2 = 48 - R_1\). Substitute into equation (2): \(\frac{1}{R_1} + \frac{1}{48-R_1} = \frac{1}{9}\). Clear denominators and solve the quadratic: \(9(R_1 + (48-R_1)) = R_1(48-R_1)\).After simplification: \(R_1^2 - 48R_1 + 432 = 0\). Solve using the quadratic formula: \(R_1 = \frac{48 \pm \sqrt{48^2 - 4 imes 432}}{2}\). This gives \(R_1 = 36\; \Omega\) and \(R_2 = 12\; \Omega\).

Analyze Power Consumption in Series

In series, power is distributed according to \( P = I^2R \). The total current \( I \) is \(\frac{48}{48}\). Thus, \( P_1 = I^2 \times R_1 = \frac{36}{48} \times 48 = 36\; W\) and \( P_2 = 12\; W\). So, \(R_1\) consumes more power in series.

Analyze Power Consumption in Parallel

In parallel, the voltage across each resistor is the same (48 V). Power is distributed as \( P = \frac{V^2}{R} \). Therefore, \( P_1 = \frac{48^2}{36} = 64\; W\) and \( P_2 = \frac{48^2}{12} = 192\; W\). Thus, \(R_2\) consumes more power in parallel.

Key concepts

Ohm's Law

Ohm’s Law is a foundational principle in electronics that describes the relationship between voltage, current, and resistance in an electrical circuit. It is represented by the equation \( V = IR \), where \( V \) is the voltage across the component, \( I \) is the current flowing through it, and \( R \) is the resistance. This law is incredibly useful for calculating the unknown quantities in a circuit when the other two are known.

For example, if you know the current flowing through a resistor and the voltage across it, you can determine its resistance. Similarly, knowing the resistance and the voltage allows you to calculate the current. In practical scenarios, like the one in the exercise, Ohm's Law helps in determining the resistance values when combined with formulas for power calculations.

Ohm’s Law is universally applied in both simple and complex circuits, providing a straightforward approach to understanding and solving electrical problems. Its simplicity and universal application make it an essential tool for anyone working with electronics.

Power in Series Circuits

In series circuits, the current flowing through each component is the same because there is only one path for the flow of electric charge. The total resistance in a series circuit is the sum of individual resistances, represented by \( R_s = R_1 + R_2 + \ldots \). Therefore, in a series circuit with resistors, the total resistance increases as more resistors are added.

Power in a series circuit is distributed among the resistors based on their resistances. The power consumed by each resistor can be calculated using the formula \( P = I^2 R \), where \( I \) is the total current, which remains constant across the circuit.

In the exercise, for the resistors in series, the current can be found using \( I = \frac{V}{R_s} \), where \( V \) is the voltage of the battery. Each resistor consumes power proportionally to its resistance. Consequently, as demonstrated in the solution, the resistor with the higher resistance (\( R_1 = 36 \Omega \)) consumes more power, \( 36 \text{ W} \), compared to the lower resistance (\( R_2 = 12 \Omega \)), which consumes \( 12 \text{ W}\). Understanding these power relationships in series circuits is crucial for designing efficient electronic devices.

Power in Parallel Circuits

In parallel circuits, each component is connected directly to the voltage source, meaning each resistor has the same voltage across it. This contrasts with series circuits, where only the current remains constant. In parallel, the total resistance is calculated using the reciprocal formula \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots \). The result is typically a total resistance that is lower than any individual resistance in the network.

Power distribution in parallel circuits follows the formula \( P = \frac{V^2}{R} \). Hence, the power consumed by each resistor is inversely related to its resistance since the voltage \( V \) is constant across each resistor.

In the example provided, each resistor has the same voltage of 48 V. Therefore, the power consumed by \( R_1 = 36 \Omega \) is \( 64 \text{ W} \), and for \( R_2 = 12 \Omega \), it is \( 192 \text{ W} \). The lower resistance consumes more power, a typical characteristic in parallel circuits. This concept highlights why careful distribution and management of power are necessary in electronic design to ensure components operate safely and efficiently.