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A 42-\(\Omega\) resistor and a 20\(\Omega\) resistor are connected in parallel, and the combination is connected across a 240-V dc line. (a) What is the resistance of the parallel combination? (b) What is the total current through the parallel combination? (c) What is the current through each resistor?

Short Answer

Expert verified
(a) 13.55 Ω, (b) 17.71 A, (c) 5.71 A through 42 Ω and 12.0 A through 20 Ω.

Step by step solution

01

Identify the formula for parallel resistance

For resistors in parallel, the equivalent resistance can be found using the formula: \( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \), where \( R_1 = 42 \Omega \) and \( R_2 = 20 \Omega \).
02

Calculate equivalent resistance

Substitute the resistor values into the formula: \( \frac{1}{R_{eq}} = \frac{1}{42} + \frac{1}{20} \). Calculate \( \frac{1}{R_{eq}} \) to get \( \frac{1}{R_{eq}} = \frac{0.0238 + 0.05}{1} = 0.0738 \). Therefore, \( R_{eq} = \frac{1}{0.0738} \approx 13.55 \Omega \).
03

Use Ohm's Law to find total current

The total current through the parallel combination can be found using Ohm's Law, \( I_{total} = \frac{V}{R_{eq}} \). Substitute \( V = 240 \) V and \( R_{eq} = 13.55 \Omega \) into the equation to get \( I_{total} = \frac{240}{13.55} \approx 17.71 \) A.
04

Calculate individual currents

For resistors in parallel, the voltage across each resistor is the same. Calculate the current through each resistor using Ohm's Law. For the 42 \( \Omega \) resistor, \( I_1 = \frac{240}{42} \approx 5.71 \) A. For the 20 \( \Omega \) resistor, \( I_2 = \frac{240}{20} = 12.0 \) A.

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Key Concepts

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

Equivalent Resistance
When dealing with electrical circuits, especially those having resistors in parallel, understanding equivalent resistance is crucial. Equivalent resistance is a way to simplify complex circuits by reducing multiple resistors into a single one. This simplifies the analysis of the circuit. When resistors are connected in parallel, they share the same voltage but can carry different currents.
The formula to calculate the equivalent resistance for resistors in parallel is:
  • \[\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n}.\]
This equation essentially states that the reciprocal of the total resistance is the sum of the reciprocals of each individual resistance.
In our example, we have two resistors, a 42-ohm (Ω) resistor and a 20-ohm resistor in parallel. When applying the formula, we get the equivalent resistance as approximately 13.55 Ω. This means that for purposes of calculating total current or analyzing the circuit further, these two resistors can be replaced with a single resistor of 13.55 Ω.
Ohm's Law
Ohm's Law is a fundamental principle in electronics that relates voltage, current, and resistance in an electrical circuit. The law is typically expressed by the equation:
  • \[V = IR,\]
where \(V\) stands for voltage, \(I\) is the current, and \(R\) represents resistance. This formula is incredibly useful for calculating the missing variable when two of these quantities are known.
For instance, in scenarios where you know the voltage across a resistor and its resistance, you can determine the current flowing through it by rearranging Ohm's Law to:
  • \[I = \frac{V}{R}.\]
In the given problem, after finding the equivalent resistance of 13.55 Ω, we employed Ohm's Law to find the total current through the parallel resistors. With a voltage of 240 V, the total current was calculated to be about 17.71 Amperes (A).
This current is then distributed among the parallel resistors, taking into account their equal voltage drops.
Electrical Circuits
Electrical circuits are paths which allow electric charges to move and power various devices. A circuit typically consists of a power source, like a battery, conductors (such as wires), and various components like resistors. Resistors in a circuit control the flow of electrical current by restricting it, which is crucial in protecting delicate components from receiving too much current.
Circuits can be configured in a variety of ways, with parallel and series connections being the most common. In a parallel circuit configuration, components are connected across the same two points, sharing the voltage. Each component is independently connected to the power supply. Therefore, if one path is broken, the current can still travel through other paths.
In our specific case, we examined two resistors arranged in parallel. The total circuit included a 240-V dc line providing the necessary voltage. By understanding the behavior of current and voltage in this parallel configuration, we've been able to calculate the current for each resistor using simple principles of circuit analysis and Ohm's Law, demonstrating fundamental circuit interactions.

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Most popular questions from this chapter

The electronics supply company where you work has two different resistors, \(R_1\) and \(R_2\), in its inventory, and you must measure the values of their resistances. Unfortunately, stock is low, and all you have are \(R_1\) and \(R_2\) in parallel and in series-and you can't separate these two resistor combinations. You separately connect each resistor network to a battery with emf 48.0 V and negligible internal resistance and measure the power \(P\) supplied by the battery in both cases. For the series combination, \(P =\) 48.0 W; for the parallel combination, \(P =\) 256 W. You are told that \(R_1\) \(R_2\). (a) Calculate \(R_1\) and \(R_2\). (b) For the series combination, which resistor consumes more power, or do they consume the same power? Explain. (c) For the parallel combination, which resistor consumes more power, or do they consume the same power?

Assume that a typical open ion channel spanning an axon's membrane has a resistance of 1 \(\times\) 10\(^{11}\) \(\Omega\). We can model this ion channel, with its pore, as a 12-nm-long cylinder of radius 0.3 nm. What is the resistivity of the fluid in the pore? (a) 10 \(\Omega\) \(\cdot\) m; (b) 6 \(\Omega\) \(\cdot\) m; (c) 2 \(\Omega\) \(\cdot\) m; (d) 1 \(\Omega\) \(\cdot\) m.

A resistor with \(R =\) 850 \(\Omega\) is connected to the plates of a charged capacitor with capacitance \(C =\) 4.62 \(\mu\)F. Just before the connection is made, the charge on the capacitor is 6.90 mC. (a) What is the energy initially stored in the capacitor? (b) What is the electrical power dissipated in the resistor just after the connection is made? (c) What is the electrical power dissipated in the resistor at the instant when the energy stored in the capacitor has decreased to half the value calculated in part (a)?

Three resistors having resistances of 1.60 \(\Omega\), 2.40 \(\Omega\), and 4.80 \(\Omega\) are connected in parallel to a 28.0-V battery that has negligible internal resistance. Find (a) the equivalent resistance of the combination; (b) the current in each resistor; (c) the total current through the battery; (d) the voltage across each resistor; (e) the power dissipated in each resistor. (f) Which resistor dissipates the most power: the one with the greatest resistance or the least resistance? Explain why this should be.

A 12.4-\(\mu\)F capacitor is connected through a 0.895-M\(\Omega\) resistor to a constant potential difference of 60.0 V. (a) Compute the charge on the capacitor at the following times after the connections are made: 0, 5.0 s, 10.0 s, 20.0 s, and 100.0 s. (b) Compute the charging currents at the same instants. (c) Graph the results of parts (a) and (b) for \(t\) between 0 and 20 s.

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