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Chapter 25: Problem 2

You make a parallel combination of resistors consisting of 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 the resistor A. b) slightly less than the resistance of the resistor \(\mathrm{A}\). c) slightly greater than the resistance of the resistor B. d) slightly less than the resistance of the resistor B.

Short Answer

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Question: When a resistor A has a very large resistance (almost infinity) and a resistor B has a very small resistance (almost zero) are connected in parallel, the equivalent resistance of the parallel resistors will be _____: a) equal to the resistance of the resistor A b) equal to the resistance of the resistor B c) slightly greater than the resistance of the resistor B d) slightly smaller than the resistance of the resistor A Answer: c) slightly greater than the resistance of the resistor B

Step by step solution

01

Formula for parallel resistors

Recall the formula for calculating the equivalent resistance when two resistors are in parallel: \(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2}\)
02

Understand the problem

We are given that resistor A has a very large resistance (almost infinity) and resistor B has a very small resistance (almost zero). Let's consider their respective resistances as \(R_A\) and \(R_B\).
03

Apply the formula

Apply the formula for equivalent resistance with the given values: \(R_{eq} = \frac{R_A \times R_B}{R_A + R_B}\)
04

Analyze the result

Considering that \(R_A\) is very large and \(R_B\) is very small, the denominator (\(R_A + R_B\)) will be very close to \(R_A\) and the numerator (\(R_A \times R_B\)) will be a very small value compared to \(R_A\). Therefore, when divided, the equivalent resistance will be slightly greater than the very small resistance \(R_B\). Thus, the correct answer is: c) slightly greater than the resistance of the resistor B.

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Most popular questions from this chapter

A potential difference of \(V=0.500 \mathrm{~V}\) is applied across a block of silicon with resistivity \(8.70 \cdot 10^{-4} \Omega \mathrm{m}\). As indicated in the figure, the dimensions of the silicon block are width \(a=2.00 \mathrm{~mm}\) and length \(L=15.0 \mathrm{~cm} .\) The resistance of the silicon block is \(50.0 \Omega\), and the density of charge carriers is \(1.23 \cdot 10^{23} \mathrm{~m}^{-3}\) Assume that the current density in the block is uniform and that current flows in silicon according to Ohm's Law. The total length of 0.500 -mm-diameter copper wire in the circuit is \(75.0 \mathrm{~cm},\) and the resistivity of copper is \(1.69 \cdot 10^{-8} \Omega \mathrm{m}\) a) What is the resistance, \(R_{w}\) of the copper wire? b) What are the direction and the magnitude of the electric current, \(i\), in the block? c) What is the thickness, \(b\), of the block? d) On average, how long does it take an electron to pass from one end of the block to the other? \(?\) e) How much power, \(P\), is dissipated by the block? f) In what form of energy does this dissipated power appear?

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