Question

You make a parallel connection between two resistors, resistor \(A\) having a very large resistance and resistor B having a very small resistance. The equivalent resistance for this combination will be a) slightly greater than the resistance of resistor A. b) slightly less than the resistance of resistor A. c) slightly greater than the resistance of resistor B. d) slightly less than the resistance of resistor B.

Short answer

a) slightly greater than the resistance of resistor A. b) slightly less than the resistance of resistor A. c) slightly greater than the resistance of resistor B. d) slightly less than the resistance of resistor B. Answer: b) slightly less than the resistance of resistor A, and c) slightly greater than the resistance of resistor B.

Step-by-step solution

Understand the parallel combination formula for resistors

When resistors are connected in parallel, the equivalent resistance is calculated using the following formula: \(\frac{1}{R_{eq}} = \frac{1}{R_A} + \frac{1}{R_B}\), where \(R_{eq}\) is the equivalent resistance, \(R_A\) is the resistance of resistor A, and \(R_B\) is the resistance of resistor B.

Obtain the relationship between equivalent resistance and each resistor

We want to analyze how the equivalent resistance will be related to the resistances of each resistor. Let's use the given information and rewrite the equation: Given resistor A has a very large resistance, we can call it as \(R_A = L\), where L is a very large number. Similarly, given resistor B has a very small resistance, we can call it as \(R_B = S\), where S is a very small number. Now, rewrite the equation for the equivalent resistance as: \(\frac{1}{R_{eq}} = \frac{1}{L} + \frac{1}{S}\)

Analyze the equivalent resistance

We know that for any two positive numbers, the sum of their reciprocals will always be greater than the reciprocal of the larger number, i.e., \(\frac{1}{L} + \frac{1}{S} > \frac{1}{L}\) Now, let's compare one divided by the sum of the reciprocals to the reciprocals of each resistor: Since \(R_{eq} = \frac{1}{\frac{1}{L} + \frac{1}{S}}\), - If we compare with the reciprocal of resistor A, we get \(\frac{1}{R_{eq}} > \frac{1}{L}\) which means \(R_{eq} < L\). - If we compare with the reciprocal of resistor B, we get \(\frac{1}{R_{eq}} < \frac{1}{S}\) which means \(R_{eq} > S\). Thus, we find that the equivalent resistance is less than the resistance of resistor A and greater than the resistance of resistor B. Comparing this to the given options, a) slightly greater than the resistance of resistor A. (Incorrect) b) slightly less than the resistance of resistor A. (Correct) c) slightly greater than the resistance of resistor B. (Correct) d) slightly less than the resistance of resistor B. (Incorrect) So, the correct answer is option b) slightly less than the resistance of resistor A, and option c) slightly greater than the resistance of resistor B.