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Chapter 19: Problem 47

Calculate the (a) maximum and (b) minimum values of resistance that can be obtained by combining resistors of \(36 \Omega,\) \(47 \Omega,\) and 51\(\Omega .\)

Short Answer

Expert verified
Max resistance: 134 惟 in series; Min resistance: ~14.56 惟 in parallel.

Step by step solution

01

Identify Possible Combinations

To find the maximum and minimum resistance values, we need to consider the possible combinations of arranging the resistors: all in series, all in parallel, or combinations of both.
02

Calculate Maximum Resistance (Series Combination)

In a series combination, the total resistance is simply the sum of all resistor values. So, calculate:\[R_{max} = R_1 + R_2 + R_3 = 36 \, \Omega + 47 \, \Omega + 51 \, \Omega\]Calculate the sum to find the maximum resistance.
03

Solution to Series Calculation

Sum the resistors:\[ R_{max} = 36 + 47 + 51 = 134 \, \Omega \]Therefore, the maximum resistance obtainable by connecting the resistors in series is 134 惟.
04

Calculate Minimum Resistance (Parallel Combination)

For resistors in parallel, the reciprocal of the total resistance is the sum of the reciprocals of each resistance:\[\frac{1}{R_{min}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\]Substitute the known values:\[\frac{1}{R_{min}} = \frac{1}{36} + \frac{1}{47} + \frac{1}{51}\]Calculate the values to find the reciprocal of the minimum resistance.
05

Solution to Parallel Calculation

Perform the calculations:\[\frac{1}{R_{min}} \approx \frac{1}{36} + \frac{1}{47} + \frac{1}{51} \approx 0.0278 + 0.0213 + 0.0196 \approx 0.0687\]Therefore, the minimum resistance is:\[ R_{min} = \frac{1}{0.0687} \approx 14.56 \, \Omega \]
06

Conclusion

By arranging the resistors in series, the maximum resistance obtained is 134 惟. By arranging them in parallel, the minimum resistance obtained is approximately 14.56 惟.

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Key Concepts

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

Understanding Series Resistance
When resistors are connected in a series, they are aligned end-to-end so that the current flows through each resistor one after the other. The key principle here is that the total resistance of the series is simply the sum of each individual resistor's resistance. This is because each additional resistor adds more opposition to the flow of electric current.

Series resistance can be calculated using the following formula:
  • \[ R_{total} = R_1 + R_2 + R_3 + \ldots \]
In our original exercise, the resistors with values 36 惟, 47 惟, and 51 惟 are combined in series to achieve the maximum possible resistance. By adding these values together, we find that the total resistance is \( 134 \, \Omega \), meaning the circuit resists the current flow the most when arranged in this way.
Understanding this concept makes it easier to design circuits where high resistance is required, such as in limiting current flow to protect sensitive components.
Exploring Parallel Resistance
In a parallel circuit, resistors are connected across the same two points, providing multiple paths for the electrical charge to flow. Here, the total resistance decreases as more resistors are added. This is because a parallel arrangement allows for more pathways for the current, reducing the load on each resistor.

The formula used to calculate total resistance in parallel is the reciprocal of the sum of the reciprocals of each resistor:
  • \[ \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots \]
In our exercise, by applying this formula to resistors 36 惟, 47 惟, and 51 惟, we calculated the total resistance to be approximately \( 14.56 \, \Omega \). This shows the minimum resistance because multiple paths allow for more current to pass through the circuit.
Parallel resistances are useful in applications where low resistance is desired, such as in ensuring power is supplied efficiently over long distances.
Performing Resistance Calculations
Resistance calculations are fundamental in electrical engineering to determine how resistors affect a circuit. Calculating resistance accurately is crucial for designing circuits that function correctly and safely.

When calculating resistance:
  • For series resistors, simply sum them up using: \( R_{total} = R_1 + R_2 + \ldots \)
  • For parallel resistors, use the reciprocal formula: \( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots \)
These calculations allow you to foresee how changes in resistor configurations will influence the overall resistance of the circuit. Accurate calculations are vital for ensuring each component receives the intended voltage and current, protecting the devices from potential damage.
Whether designing a simple circuit or a robust electrical system, understanding how to perform these basic resistance calculations will help improve your designs and enhance reliability.

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Most popular questions from this chapter

A 1.50 -m cylindrical rod of diameter 0.500 \(\mathrm{cm}\) is connected to a power supply that maintains a constant potential difference of 15.0 \(\mathrm{V}\) across its ends, while an ammeter measures the current through it. You observe that at room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) the ammeter reads 18.5 \(\mathrm{A}\) , while at \(92.0^{\circ} \mathrm{C}\) it reads 17.2 \(\mathrm{A} .\) You can ignore any thermal expansion of the rod. Find (a) the resistivity and (b) the temperature coefficient of resistivity at \(20^{\circ} \mathrm{C}\) for the material of the rod.

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?

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?)

A refrigerator draws 3.5 \(\mathrm{A}\) of current while operating on a 120 \(\mathrm{V}\) power line. If the refrigerator runs 50\(\%\) of the time and electric power costs \(\$ 0.12\) per kWh, how much does it cost to run this refrigerator for a 30 -day month?

Electric eels. Electric eels generate electric pulses along their skin that can be used to stun an enemy when they come into contact with it. Tests have shown that these pulses can be up to 500 \(\mathrm{V}\) and produce currents of 80 \(\mathrm{mA}\) (or even larger). A typical pulse lasts for 10 \(\mathrm{ms}\) . What power and how much energy are delivered to the unfortunate enemy with a single pulse, assuming a steady current?
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