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Chapter 25: Problem 16

Two resistors with resistances \(R_{1}\) and \(R_{2}\) are connected in parallel. Demonstrate that, no matter what the actual values of \(R_{1}\) and \(R_{2}\) are, the equivalent resistance is always less than the smaller of the two resistances.

Short Answer

Expert verified
Short Answer: The equivalent resistance of two resistors connected in parallel is always less than the smallest individual resistance. This is demonstrated by the formula \(R_{eq} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}\) which, after analysis, shows that the equivalent resistance is smaller than the larger of the two resistances and, therefore, guaranteed to be smaller than the smallest resistance.

Step by step solution

01

Parallel Resistance Formula

To find the equivalent resistance (\(R_{eq}\)) of two resistors with resistances \(R_{1}\) and \(R_{2}\) connected in parallel, we use the formula: \[ \frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \]
02

Derive the Formula to find \(R_{eq}\)

Now, we'll derive the formula from Step 1 to find \(R_{eq}\): \[ R_{eq} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} \]
03

Analyze the Formula

To demonstrate that the equivalent resistance is always less than the smaller of the two resistances, we can analyze the formula: Let the smaller resistance be \(R_s\) and the larger resistance be \(R_l\), where \(R_s<R_l\). Since the numerator is the product of the two resistances (\(R_{1} R_{2}\)), and the denominators (\(R_{1} + R_{2}\)) will always be greater than the smaller resistance (\(R_s\)), it means: \[ \frac{R_s R_l}{R_s + R_l} < \frac{R_s R_l}{R_s} \]
04

Simplify the Expression

Now, let's simplify the expression obtained in Step 3: \[ R_{eq} < \frac{R_s R_l}{R_s} \] Cancel out \(R_s\) from numerator and denominator: \[ R_{eq} < R_l \] Since \(R_l > R_s\), we have now shown that \(R_{eq}\) is smaller than the larger resistance (\(R_l\)), which means it is guaranteed to be smaller than the smallest resistance (\(R_s\)). Thus, the equivalent resistance is always less than the smaller of the two individual resistances.

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Most popular questions from this chapter

A 34 -gauge copper wire, with a constant potential difference of \(0.10 \mathrm{~V}\) applied across its \(1.0 \mathrm{~m}\) length at room temperature \(\left(20 .{ }^{\circ} \mathrm{C}\right),\) is cooled to liquid nitrogen temperature \(\left(77 \mathrm{~K}=-196^{\circ} \mathrm{C}\right)\) a) Determine the percentage change in the wire's resistance during the drop in temperature. b) Determine the percentage change in current flowing in the wire. c) Compare the drift speeds of the electrons at the two temperatures.

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A copper wire has a diameter \(d_{\mathrm{Cu}}=0.0500 \mathrm{~cm}\) is \(3.00 \mathrm{~m}\) long, and has a density of charge carriers of \(8.50 \cdot 10^{28}\) electrons \(/ \mathrm{m}^{3}\). As shown in the figure, the copper wire is attached to an equal length of aluminum wire with a diameter \(d_{\mathrm{A} \mathrm{I}}=0.0100 \mathrm{~cm}\) and density of charge carriers of \(6.02 \cdot 10^{28}\) electrons \(/ \mathrm{m}^{3}\). A current of 0.400 A flows through the copper wire. a) What is the ratio of the current densities in the two wires, \(J_{\mathrm{Cu}} / J_{\mathrm{Al}} ?\) b) What is the ratio of the drift velocities in the two wires, \(v_{\mathrm{d}-\mathrm{Cu}} / v_{\mathrm{d}-\mathrm{Al}} ?\)

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