Question

Resistors in parallel Two resistors in an electrical circuit with resistance \(R_{1}\) and \(R_{2}\) wired in parallel with a constant voltage give an effective resistance of \(R,\) where \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\). (Check your book to see figure) a. Find \(\frac{\partial R}{\partial R_{1}}\) and \(\frac{\partial R}{\partial R_{2}}\) by solving for \(R\) and differentiating. b. Find \(\frac{\partial R}{\partial R_{1}}\) and \(\frac{\partial R}{\partial R_{2}}\) by differentiating implicitly. c. Describe how an increase in \(R_{1}\) with \(R_{2}\) constant affects \(R\) d. Describe how a decrease in \(R_{2}\) with \(R_{1}\) constant affects \(R\)

Short answer

Answer: When R_1 increases while R_2 remains constant, the effective resistance R increases. When R_2 decreases while R_1 remains constant, the effective resistance R decreases.

Step-by-step solution

Part a: Solving for R and differentiating explicitly

To differentiate \(R\) with respect to \(R_1\) and \(R_2\), we first need to solve the equation \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\) for \(R\). 1. Find a common denominator for the right-hand side of the equation: \(\frac{1}{R}=\frac{R_{1}+R_{2}}{R_{1}R_{2}}\) 2. Invert both sides of the equation to get \(R\) by itself: \(R = \frac{R_{1}R_{2}}{R_{1}+R_{2}}\) Now we can take the partial derivatives with respect to \(R_1\) and \(R_2\):

Part a: Derivative with respect to R_1

Differentiate \(\frac{R_{1}R_{2}}{R_{1}+R_{2}}\) with respect to \(R_1\) using the quotient rule: \(\frac{\partial R}{\partial R_{1}} = \frac{(R_{1} + R_{2})(R_{2}) - (R_{1}R_{2})(1)}{(R_{1} + R_{2})^2} = \frac{R_{2}^2}{(R_{1}+R_{2})^2}\)

Part a: Derivative with respect to R_2

Differentiate \(\frac{R_{1}R_{2}}{R_{1}+R_{2}}\) with respect to \(R_2\) using the quotient rule: \(\frac{\partial R}{\partial R_{2}} = \frac{(R_{1} + R_{2})(R_{1}) - (R_{1}R_{2})(1)}{(R_{1} + R_{2})^2} = \frac{R_{1}^2}{(R_{1}+R_{2})^2}\)

Part b: Differentiating implicitly

Now let's find the partial derivatives of R with respect to \(R_{1}\) and \(R_{2}\) by differentiating implicitly. We start with the original equation \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\).

Part b: Implicit derivative with respect to R_1

Differentiate both sides of the equation with respect to \(R_1\), treating \(R, R_1, R_2\) as functions of \(R_1\), and solve for \(-\frac{\partial R}{\partial R_1}\): \(\frac{-\partial R/\partial R_1}{R^2} = \frac{-1}{R_1^2} \Rightarrow \frac{\partial R}{\partial R_1} = \frac{R^2}{R_1^2}\) Replace \(R\) with the expression found in part a: \(\frac{\partial R}{\partial R_1} = \frac{(R_1R_2/(R_1+R_2))^2}{R_1^2} = \frac{R_2^2}{(R_1+R_2)^2}\)

Part b: Implicit derivative with respect to R_2

Differentiate both sides of the equation with respect to \(R_2\), treating \(R, R_1, R_2\) as functions of \(R_2\), and solve for \(-\frac{\partial R}{\partial R_2}\): \(\frac{-\partial R/\partial R_2}{R^2} = \frac{-1}{R_2^2} \Rightarrow \frac{\partial R}{\partial R_2} = \frac{R^2}{R_2^2}\) Replace \(R\) with the expression found in part a: \(\frac{\partial R}{\partial R_2} = \frac{(R_1R_2/(R_1+R_2))^2}{R_2^2} = \frac{R_1^2}{(R_1+R_2)^2}\) Since the results match with those from part a, we have successfully found the partial derivatives by using implicit differentiation.

c. Effect on R of increasing R_1

Now we will examine how an increase in \(R_1\) with \(R_2\) constant affects \(R\). We can analyze this based on the partial derivative with respect to \(R_1\), which is: \(\frac{\partial R}{\partial R_1} = \frac{R_2^2}{(R_1+R_2)^2}\) This expression is positive (assuming \(R_1\) and \(R_2\) are positive), which means that as \(R_1\) increases, \(R\) also increases. Thus, if \(R_1\) increases while \(R_2\) remains constant, the effective resistance \(R\) increases.

d. Effect on R of decreasing R_2

Lastly, we will examine how a decrease in \(R_2\) with \(R_1\) constant affects \(R\). We can analyze this based on the partial derivative with respect to \(R_2\): \(\frac{\partial R}{\partial R_2} = \frac{R_1^2}{(R_1+R_2)^2}\) This expression is positive (assuming \(R_1\) and \(R_2\) are positive), which means that as \(R_2\) increases, \(R\) also increases. Thus, if \(R_2\) decreases while \(R_1\) remains constant, the effective resistance \(R\) will also decrease (the opposite effect of increasing \(R_2\)).

Key concepts

Implicit Differentiation

Implicit differentiation is a powerful technique when dealing with equations where variables are tangled, and it is challenging to solve for one variable in terms of others. Instead of isolating one variable, we differentiate everything with respect to the variable of interest.
This technique is handy when we have a function defined by an equation, such as \[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \]. Here, it's more complex to express R directly in terms of \(R_1\) or \(R_2\), so we differentiate all parts of the equation.

• To apply implicit differentiation here, differentiate both sides of the given equation with respect to \(R_1\) or \(R_2\), treating every variable that isn't constant as a function.
• Remember to apply the chain rule when necessary. For example, the derivative of \( \frac{1}{R_1} \) with respect to \(R_1\) is \( -\frac{1}{R_1^2} \).
• After differentiating, solve for the partial derivatives, such as \( \frac{\partial R}{\partial R_1} \), which provide insight into how the effective resistance \(R\) changes.

Through implicit differentiation, you can still find how changes in one resistor affect the overall resistance without rearranging the formula for \(R\).

Quotient Rule in Calculus

The quotient rule is used to differentiate functions that are given as one function divided by another. If you have a function in the form \( f(x) = \frac{u(x)}{v(x)} \), the derivative is given by:
\[ f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \]
where \(u'(x)\) and \(v'(x)\) are derivatives of \(u(x)\) and \(v(x)\).
The exercise to find \( \frac{\partial R}{\partial R_1} \) and \( \frac{\partial R}{\partial R_2} \) involves using this rule, applied to the function \( R = \frac{R_1R_2}{R_1 + R_2} \).

• For \( \frac{\partial R}{\partial R_1} \), treat \(R_2\) as constant and differentiate the function. The result would be \( \frac{R_2^2}{(R_1 + R_2)^2} \).
• For \( \frac{\partial R}{\partial R_2} \), treat \(R_1\) as constant and apply the quotient rule similarly, yielding \( \frac{R_1^2}{(R_1 + R_2)^2} \).

Understanding the quotient rule lets you systematically differentiate complex ratios, giving you insight into the relationships between varying quantities.

Parallel Resistors Formula

In circuits, resistors can be combined in different configurations to achieve desired electrical properties. When two resistors are in parallel, the effective or total resistance \(R\) can be determined using the formula:
\[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \].
This formula shows that the reciprocal of the total resistance is the sum of the reciprocals of the individual resistances.

• One key takeaway is that the total resistance of resistors in parallel is always less than the smallest individual resistance. This is because adding more paths for the current to flow actually reduces the overall resistance.
• This concept can be extended to more than two resistors, where the formula would involve adding the reciprocals of all resistances involved.

Using this formula, if you have a network of resistors, you can determine how changes in any one resistor affect the overall resistance by analyzing the partial derivatives. In the exercise, this understanding helps in predicting the effects of changing \(R_1\) or \(R_2\) on the total resistance \(R\).