Question

Two electrical resistors When two electrical resistors with resistance \(R_{1}>0\) and \(R_{2}>0\) are wired in parallel in a circuit (see figure), the combined resistance \(R,\) measured in ohms \((\Omega),\) is given by \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\) a. Estimate the change in \(R\) if \(R_{1}\) increases from \(2 \Omega\) to \(2.05 \Omega\) and \(R_{2}\) decreases from \(3 \Omega\) to \(2.95 \Omega\) b. Is it true that if \(R_{1}=R_{2}\) and \(R_{1}\) increases by the same small amount as \(R_{2}\) decreases, then \(R\) is approximately unchanged? Explain. c. Is it true that if \(R_{1}\) and \(R_{2}\) increase, then \(R\) increases? Explain. d. Suppose \(R_{1}>R_{2}\) and \(R_{1}\) increases by the same small amount as \(R_{2}\) decreases. Does \(R\) increase or decrease?

Short answer

Answer: In this scenario, the combined resistance R will likely decrease.

Step-by-step solution

Apply the formula for resistances in parallel

In order to answer all parts of this exercise, use the formula: \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\).

Calculate R for the given values in part (a)

In part (a), we know \(R_1=2\Omega\) and \(R_2=3\Omega\). Using the formula, we get: $$\frac{1}{R}=\frac{1}{2}+\frac{1}{3}$$ Solve for \(R\) to obtain the initial combined resistance: $$R=\frac{6}{5}\Omega$$

Calculate R for the new values in part (a)

Now, we consider the case where \(R_1=2.05\Omega\) and \(R_2=2.95\Omega\). Substitute these new values into the formula: $$\frac{1}{R}=\frac{1}{2.05}+\frac{1}{2.95}$$ Solve for \(R\) to get the new combined resistance: $$R\approx1.194\Omega$$

Estimate the change in R for part (a)

Find the difference between the initial combined resistance and the new combined resistance: $$\Delta R = R_{new} - R_{initial} \approx 1.194 \Omega -\frac{6}{5}\Omega \approx -0.006\Omega$$

Analyze the relationship between R1 and R2 for part (b)

When \(R_{1}=R_{2}\), the formula becomes: $$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{1}}\Rightarrow R=\frac{R_{1}}{2}$$ If \(R_1\) increases by the same small amount as \(R_2\) decreases, the change of combined resistance \(\Delta R\) will be very close to zero and \(R\) will be approximately unchanged.

Analyze the relationship between increased resistances for part (c)

For part (c), we need to show if R increases when both \(R_1\) and \(R_2\) increase. We can use the concept of partial derivatives. The partial derivatives must be negative for the combined resistance to decrease when the individual resistances increase. $$\frac{\partial}{\partial R_1}\left(\frac{1}{R}\right) = - \frac{1}{R_1^2} < 0$$ $$\frac{\partial}{\partial R_2}\left(\frac{1}{R}\right) = - \frac{1}{R_2^2} < 0$$ Since both partial derivatives are negative, if \(R_1\) and \(R_2\) increase, \(R\) will decrease.

Determine whether R increases or decreases for part (d)

For part (d), we have \(R_1 > R_2\) and \(R_1\) increases by the same small amount as \(R_2\) decreases. We can analyze this scenario by computing the total differential of \(\frac{1}{R}\) with respect to \(R_1\) and \(R_2\). $$d\left(\frac{1}{R}\right) = \frac{\partial}{\partial R_1}\left(\frac{1}{R}\right)dR_1 + \frac{\partial}{\partial R_2}\left(\frac{1}{R}\right)dR_2$$ Since \(R_1\) increases and \(R_2\) decreases by the same small amount, \(dR_1 = - dR_2\). Substitute this into the total differential equation: $$d\left(\frac{1}{R}\right) = -\frac{1}{R_1^2}dR_1 - \frac{1}{R_2^2}(-dR_1) = dR_1\left(\frac{1}{R_2^2} - \frac{1}{R_1^2}\right)$$ Because \(R_1 > R_2\), the term \((\frac{1}{R_2^2} - \frac{1}{R_1^2}) > 0\), and \(d\left(\frac{1}{R}\right)\) is positive. So, when \(R_1\) increases by the same small amount as \(R_2\) decreases, the combined resistance \(R\) will decrease.

Key concepts

Electrical Resistance

Electrical resistance is a fundamental concept when discussing electricity as it relates to how strongly a component opposes the flow of electrical current. The unit of measurement for resistance is the ohm, symbolized as \(\Omega\). Whenever we place resistors in parallel within a circuit, the overall or combined resistance decreases. This is because the electrical current has multiple paths to travel through, reducing the total opposition.

For two resistors with resistances \(R_1\) and \(R_2\), the formula for their combined resistance \(R\) in parallel is:\[\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\]Applying this formula allows us to analyze how variations in \(R_1\) and \(R_2\) affect the overall resistance of a circuit. This principle is vital in designing electronic systems that require specific current levels.

Partial Derivatives

In calculus, partial derivatives help us understand how a function changes as one of its variables changes, while keeping the others constant. For our resistance in parallel, partial derivatives are useful in determining how a small change in one resistor’s value affects the combined resistance.

Let's denote our combined resistance formula again:\[\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\]The partial derivative of \(\frac{1}{R}\) with respect to \(R_1\) is:\[\frac{\partial}{\partial R_1}\left(\frac{1}{R}\right) = -\frac{1}{R_1^2} < 0\]Similarly, for \(R_2\) we have:\[\frac{\partial}{\partial R_2}\left(\frac{1}{R}\right) = -\frac{1}{R_2^2} < 0\]These derivatives show that increasing either resistor individually will lead to a decrease in the combined resistance. Understanding these details can provide insightful perspectives on how electronic components interact in a network.

Change in Resistance

When resistors in a parallel arrangement experience changes in value, it mathematically alters the combined resistance of the system. Using calculus, specifically differentials, we can estimate these changes precisely.

Let's consider an initial scenario where \(R_1 = 2\Omega\) and \(R_2 = 3\Omega\). After a change, \(R_1\) becomes \(2.05\Omega\) and \(R_2\) adjusts to \(2.95\Omega\). Calculating the resistance, both initially and after change, shows us how those variations affect \(R\).

The initial combined resistance is \(\frac{6}{5}\Omega\). After the new values are applied, using our parallel formula gives us approximately \(1.194\Omega\). The change, calculated as:\[\Delta R = R_{new} - R_{initial} \approx 1.194 \Omega - \frac{6}{5} \Omega \approx -0.006 \Omega\]This change indicates a slight decrease, refined through understanding how small variations influence the entire circuit.

Mathematical Analysis

Mathematical analysis, involving functions, derivatives, and equations, provides a framework for predicting and understanding changes in physical systems like electrical circuits. Using previous examples with resistors, we can utilize total differentials to predict whether resistance increases or decreases under specific conditions.

For instance, consider a condition where \(R_1 > R_2\) and it is increased by the same amount that \(R_2\) is decreased. Calculating the impact involves the total differential, which supplants calculus into electronics:\[d\left(\frac{1}{R}\right) = \frac{\partial}{\partial R_1}\left(\frac{1}{R}\right)dR_1 + \frac{\partial}{\partial R_2}\left(\frac{1}{R}\right)dR_2\]Substituting \(dR_1 = -dR_2\), we find the result reflects the potential increase or decrease in \( \frac{1}{R} \). Because \(R_1 > R_2\), this differential is positive, making \( \frac{1}{R} \) increase, leading to \(R\) decreasing.

This analysis is a powerful tool in engineering applications, enabling precise control and prediction of system behavior.