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Chapter 19: Problem 55

You absentmindedly solder a 69.8 \(\mathrm{k} \Omega\) resistor into a circuit where a 36.5 \(\mathrm{k} \Omega\) should be. How can you get the proper resistance without replacing the bigger resistor or removing anything from the circuit?

Short Answer

Expert verified
Add a 76.52 k惟 resistor in parallel with the 69.8 k惟 resistor.

Step by step solution

01

Understanding the Problem

You have mistakenly used a 69.8 k惟 resistor instead of a 36.5 k惟 resistor. You need to adjust the resistance without removing the 69.8 k惟 resistor or adding extra components to the setup.
02

Consider Parallel Resistance

To reduce the effective resistance to 36.5 k惟, consider adding another resistor in parallel with the 69.8 k惟. This approach allows the total resistance to decrease.
03

Use Parallel Resistance Formula

For resistors in parallel, the formula is \( R_{ ext{total}} = \frac{R_1 \times R_2}{R_1 + R_2} \). Here, \( R_1 = 69.8 \) k惟 and you want \( R_{ ext{total}} = 36.5 \) k惟. You need to solve for \( R_2 \).
04

Set Up the Equation

Set the equation using the values: \[ 36.5 = \frac{69.8 \times R_2}{69.8 + R_2} \]. Rearrange this to solve for \( R_2 \).
05

Solve for R2

Rearrange the equation: \( 36.5 (69.8 + R_2) = 69.8 R_2 \), which simplifies to \( 36.5 \times 69.8 + 36.5 R_2 = 69.8 R_2 \). Simplify further to \( 36.5 \times 69.8 = (69.8 - 36.5) R_2 \).
06

Calculate the Result

Calculate \( R_2 \): \( R_2 = \frac{36.5 \times 69.8}{69.8 - 36.5} \). Doing the math: \( 36.5 \times 69.8 = 2548.7 \) and \( 69.8 - 36.5 = 33.3 \). Then, \( R_2 = \frac{2548.7}{33.3} \approx 76.52 \) k惟.

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Key Concepts

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

Parallel Resistance
Understanding parallel resistance is crucial when working with circuits. When resistors are connected in parallel, it means that they are connected across the same two points and share the same voltage. The effective resistance of resistors in parallel is different than that of resistors placed in series.
In a parallel configuration, the total resistance is lower than the smallest resistor in the group. This happens because the current has multiple paths to take, effectively decreasing the overall opposition, or resistance, in the circuit.
Key Points about Parallel Resistance:
  • Parallel resistors offer alternate paths for current flow.
  • The total equivalent resistance is less than any individual resistor in the setup.
  • The formula used is: \( R_{total} = \frac{R_1 \times R_2}{R_1 + R_2}\)
Using the formula, you can always adjust a circuit's resistance by adding new parallel resistors, thus fine-tuning your circuit's performance without removing existing components.
Circuit Troubleshooting
Circuit troubleshooting involves applying logical and systematic methods to identify and solve issues within an electrical circuit.
Whether due to human error, like in the given problem where the wrong resistor was installed, or other issues such as a damaged component.
Steps for Effective Circuit Troubleshooting:
  • Understand the circuit layout and the intended function of each component.
  • Identify and isolate the problem area within the circuit.
  • Consider potential solutions such as adjusting resistor values or adding components like parallel resistors.
  • Implement the simplest solution first and test the circuit.
  • Refine and test further if the initial solution does not fully resolve the issue.
In this exercise, adjusting the resistance with a parallel resistor offers a solution. It's important to account for how changes can affect the entire circuit; sometimes minor tweaks can significantly enhance or disrupt functionality.
Resistance Calculations
Resistance calculations are foundational to circuit design and troubleshooting. Calculating the right value ensures the circuit operates as intended.
In our specific problem, one must calculate net resistance using parallel resistance concepts to achieve a specific goal.
Step-by-Step Resistance Calculations:
  • Identify the resistors involved and their configurations within the circuit.
  • Apply the appropriate formulas, such as:\[R_{total} = \frac{R_1 \times R_2}{R_1 + R_2}\]for parallel resistors. This formula helps resolve the effective resistance issue caused by the misplaced resistor.
  • Rearrange and solve equations carefully to find unknown resistor values.
  • Check calculations several times to ensure accuracy. Even small mistakes can lead to different results, impacting the circuit's function.
  • Verify by physically testing in a non-critical circuit environment before finalizing.
Accurate resistance calculations assure that circuits deliver the desired performance, preserving both efficiency and safety.

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Most popular questions from this chapter

When a solid cylindrical rod is connected across a fixed potential difference, a current \(I\) flows through the rod. What would be the current (in terms of \(I\) ) if (a) the length were doubled, (b) the diameter were doubled, (c) both the length and the diameter were doubled?

Energy use of home appliances. An 1800 \(\mathrm{W}\) toaster, a 1400 \(\mathrm{W}\) electric frying pan, and a 75 \(\mathrm{W}\) lamp are plugged into the same electrical outlet in a \(20 \mathrm{A}, 120 \mathrm{V}\) circuit. (Note: When plugged into the same outlet, the three devices are in parallel with each other across the 120 \(\mathrm{V}\) outlet.) (a) What current is drawn by each device? (b) Will this combination blow the circuit breaker?

If we model the pore of a hypothetical ion channel spanning the membrane of a nerve cell as a cylinder 0.3 \(\mathrm{nm}\) long with a radius of 0.3 \(\mathrm{nm}\) filled with a fluid of resistivity \(100 \Omega \mathrm{cm},\) what is the conductance of the channel? (Be careful with units). A. 1 \(\mathrm{G} \Omega\) B. 1 \(\mathrm{nS}\) C. 100 \(\mathrm{G\Omega}\) D. 100 \(\mathrm{nS}\)

Navigation of electric fish. Certain fish, such as the Nile fish (Gnathonemus), concentrate charges in their head and tail, thereby producing an electric field in the water around them. (See Figure \(19.67 . )\) This field creates a potential difference of a few volts between the head and tail, which in turn causes current to flow in the conducting seawater. As the fish swims, it passes near objects that have resistivities different from that of seawater, which in turn causes the current to vary. Cells in the skin of the fish are sensitive to this current and can detect changes in it. The changes in the current allow the fish to navigate. (In the next chapter, we shall investigate how the fish might detect this current.) Since the electric field is weak far from the fish, we shall consider only the field running directly from the head to the tail. We can model the seawater through which that field passes as a conducting tube of area 1.0 \(\mathrm{cm}^{2}\) and having a potential difference of 3.0 \(\mathrm{V}\) across its ends. The length of a Nile fish is about \(20 \mathrm{cm},\) and the resistivity of seawater is 0.13\(\Omega \cdot \mathrm{m} .\) (a) How large is the current through the tube of seawater? (b) Suppose the fish swims next to an object that is 10 \(\mathrm{cm}\) long and 1.0 \(\mathrm{cm}^{2}\) in cross-sectional area and has half the resistivity of seawater. This object replaces the seawater for half the length of the tube. What is the current through the tube now? How large is the change in the current that the fish must detect? (Hint: How are this object and the remaining water in the tube connected, in series or in parallel?)

Charging and discharging a capacitor. A 1.50\(\mu \mathrm{F}\) capacitor is charged through a 125\(\Omega\) resistor and then discharged through the same resistor by short-circuiting the battery. While the capacitor is being charged, find (a) the time for the charge on its plates to reach \(1-1 / e\) of its maximum value and (b) the current in the circuit at that time. (c) During the discharge of the capacitor, find the time for the charge on its plates to decrease to 1/e of its initial value. Also, find the time for the current in the circuit to decrease to 1\(/ e\) of its initial value.
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