Question

When a \(40.0-V\) emf device is placed across two resistors in series, a current of \(10.0 \mathrm{~A}\) is flowing in each of the resistors. When the same emf device is placed across the same two resistors in parallel, the current through the emf device is \(50.0 \mathrm{~A}\). What is the magnitude of the larger of the two resistances?

Short answer

Answer: The magnitude of the larger resistor is 2.5 Ω.

Step-by-step solution

Find the equivalent resistance in series

When the resistors are in series, the equivalent resistance is the sum of their resistances. Let the resistors have resistances R1 and R2. According to Ohm's law, V = I * R. We are given the emf (V) and the current (I), so we can find the equivalent resistance in series: \(40.0 V = 10.0 A * R_{series}\) \(R_{series} = \dfrac{40.0 V}{10.0 A} = 4.0 \Omega\) As R1 and R2 are in series, \(R_{series} = R1 + R2\).

Calculate the equivalent resistance in parallel

When the resistors are in parallel, their equivalent resistance is given by: \(\dfrac{1}{R_{parallel}} = \dfrac{1}{R1} + \dfrac{1}{R2}\) The emf is still \(40.0 V\), but now we need to find the total current through the emf device, which is given as \(50.0 A\). We can use Ohm's law again to find the equivalent resistance in parallel: \(40.0 V = 50.0 A * R_{parallel}\) \(R_{parallel} = \dfrac{40.0 V}{50.0 A} = 0.8 \Omega\)

Solve for R1 and R2

Now we have two equations for R1 and R2: 1) \(R1 + R2 = 4.0 \Omega\) (Series) 2) \(\dfrac{1}{R1} + \dfrac{1}{R2} = \dfrac{1}{0.8 \Omega}\) (Parallel) To solve for R1 and R2, we can use substitution or elimination method. Let's use substitution. From the series equation, we can write R2 as: \(R2 = 4.0 \Omega - R1\) Now substitute this expression for R2 in the parallel equation: \(\dfrac{1}{R1} + \dfrac{1}{4.0 \Omega - R1} = \dfrac{1}{0.8 \Omega}\) Multiplying both sides by \(0.8 \Omega (4.0 \Omega - R1) R1\), we get: \(0.8 \Omega (4.0 \Omega - R1) + 0.8 \Omega R1 = R1 (4.0 \Omega - R1)\) Expanding and simplifying the equation, we get: \(2.4 \Omega^2 - 1.6 \Omega R1 = 0\)

Solve for R1 and determine R2

From the equation in step 3, we can factor out \(1.6 \Omega\): \(1.6 \Omega (1.5 \Omega - R1) = 0\) So, the possible solutions for R1 are: \(1.5 \Omega\) or \(R1 = 0\) (which is not physically meaningful) Therefore, R1 must be \(1.5 \Omega\). Now, from series equation: \(R2 = 4.0 \Omega - R1 = 4.0 \Omega - 1.5 \Omega = 2.5 \Omega\)

Find the magnitude of the larger resistance

We have the values of R1 and R2, which are 1.5 Ω and 2.5 Ω, respectively. The magnitude of the larger of the two resistances is 2.5 Ω.

Key concepts

Ohm's Law

Ohm's Law is a fundamental principle in the world of electronics and physics, expressing the relationship between voltage, current, and resistance in an electrical circuit. It can be stated as: Voltage (V) = Current (I) × Resistance (R). Simply put, this law implies that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, provided the temperature remains constant.

Ohm's Law is crucial when dealing with circuits, including those with resistors in series and parallel configurations. For instance, if you know the voltage across a resistor and its resistance, you can calculate the current using the formula I = V/R. Conversely, knowing the current and resistance allows you to compute the voltage. This core principle helps us to find the equivalent resistance of a series or parallel circuit by rearranging the formula to solve for the missing variable.

Understanding Ohm's Law is essential for solving a wide range of electrical problems, from simple tasks like calculating the resistance required for an LED to more complex issues like the one we see in this textbook exercise.

Equivalent Resistance

Equivalent resistance is a concept used to simplify the analysis of circuits with multiple resistors. It gives us a single value that can replace a combination of resistors without changing the overall current or voltage in the circuit.

In a series circuit, the equivalent resistance is the sum of all resistor values, since the current must pass through each resistor one after the other. The formula is R_eq-series = R₁ + R₂ + ... + R_n. This is why in our exercise, when the resistors were in series, their total resistance was simply added to find the equivalent resistance.

When resistors are placed in parallel, the equivalent resistance can be found by the formula \( 1/R_{eq-parallel} = 1/R1 + 1/R2 + ... + 1/Rn \). The total resistance in a parallel circuit is always less than the smallest individual resistance because the current has multiple paths to take, reducing the overall opposition to current flow. Hence, the exercise required us to compute the reciprocal of the sum of the reciprocals of the resistances to find the equivalent resistance in a parallel arrangement.

Electric Current

Electric current is the rate at which electric charge flows past a point in a circuit, measured in amperes (A). It results from the motion of electrons and is driven by a voltage or electromotive force (emf). Current can be either direct (DC), flowing in one direction, or alternating (AC), periodically reversing direction.

In the context of our exercise, current is what flows through the resistors when connected to an emf source. When resistors are connected in series, the current is the same through each one, as there is only one path for the current to follow. However, in a parallel arrangement, the current may split and take multiple paths, with each path carrying a fraction of the total current based on the resistance present in it. Understanding how current divides in parallel circuits and remains constant in series circuits is key to analyzing and designing electrical networks.

Electromotive Force (emf)

Electromotive force, often abbreviated as emf, is the energy provided by a cell or battery per coulomb of charge passing through it, measured in volts (V). It is essentially the voltage that drives the current in a closed circuit. An important distinction to note is that emf refers to the voltage when no current is flowing, while the actual voltage across a device when current is present may be less due to internal resistance.

In our exercise, the emf is the voltage being provided by the device across the two resistors, which remains constant at 40.0 V regardless of how the resistors are connected. The concept of emf is central to understanding how energy sources power circuits and what initiates the flow of electric current. It is especially crucial when dealing with batteries and power supplies in both theoretical problems and practical applications.