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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equivalent Resistance
When two resistors are connected in parallel, the combined resistance they offer to the flow of electric current is called the "equivalent resistance". The property of equivalent resistance helps in simplifying complex electrical circuits. To calculate the equivalent resistance of two resistors connected in parallel, we use the formula:\[ \frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \]This formula signifies that the inverse of the equivalent resistance is the sum of the inverses of each individual resistance. The resultant equivalent resistance is always lower than the smallest individual resistance in the circuit. Understanding this principle is crucial because it allows electrical engineers and enthusiasts to predict how combined resistors will behave when integrated into broader circuits. It also makes calculations manageable, leading to efficient circuit design and troubleshooting.
Resistance Formula
The resistance formula for parallel resistors is an essential concept in physics and electrical engineering. It reveals how resistors with different resistance values, when connected, combine to form a single effective resistance. The derived formula used to find the equivalent resistance \( R_{eq} \) is:\[ R_{eq} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} \]In this equation:
  • \(R_{1}\) and \(R_{2}\) are the resistances of the individual resistors.
  • The numerator is the product of the two resistances.
  • The denominator is the sum of the two resistances.
This formula indicates that the equivalent resistance is obtained by dividing the product of the resistances by their sum. Importantly, this derived equation shows that adding more parallel resistors decreases the overall resistance. This principle is leveraged in practical applications to control the current flow in circuits, making devices work safely and efficiently.
Electrical Circuits
Electrical circuits are systems composed of various electrical components like resistors, capacitors, and batteries linked by conductive paths. These components work together to control and distribute electric current to perform useful tasks. Parallel circuit designs are prevalent due to their resilience and efficiency. In a parallel configuration, multiple paths exist for current to flow. This minimises the impact when one component fails, as current can still flow through the other paths, making the circuit more reliable. Understanding how resistors in these circuits behave is crucial. When designing electronics or electrical systems:
  • Knowing the equivalent resistance helps maintain desired current levels.
  • It ensures that circuits operate within safe limits, avoiding damage to components.
  • Using parallel resistors provides flexibility in modifying circuit impedance.
By mastering equivalent resistance and parallel resistor configurations, one can create robust and efficient electrical systems that meet specific design criteria and performance needs.

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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.

A \(12.0 \mathrm{~V}\) battery with an internal resistance \(R_{\mathrm{j}}=4.00 \Omega\) is attached across an external resistor of resistance \(R\). Find the maximum power that can be delivered to the resistor.

A resistor of unknown resistance and a \(35-\Omega\) resistor are connected across a \(120-\mathrm{V}\) emf device in such a way that an 11 -A current flows. What is the value of the unknown resistance?

A charged-particle beam is used to inject a charge, \(Q_{0}\), into a small, irregularly shaped region (not a cavity, just some region within the solid block) in the interior of a block of ohmic material with conductivity \(\sigma\) and permittivity \(\epsilon\) at time \(t=0\). Eventually, all the injected charge will move to the outer surface of the block, but how quickly? a) Derive a differential equation for the charge, \(Q(t)\), in the injection region as a function of time. b) Solve the equation from part (a) to find \(Q(t)\) for all \(t \geq 0\). c) For copper, a good conductor, and for quartz (crystalline \(\mathrm{SiO}_{2}\) ), an insulator, calculate the time for the charge in the injection region to decrease by half. Look up the necessary values. Assume that the effective "dielectric constant" of copper is \(1.00000 .\)

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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