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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 \(power\) \(rating\) of a resistor is the maximum power the resistor can safely dissipate without too great a rise in temperature and hence damage to the resistor. (a) If the power rating of a 15-k \(\Omega\) resistor is 5.0 W, what is the maximum allowable potential difference across the terminals of the resistor? (b) A 9.0-k \(\Omega\) resistor is to be connected across a 120-V potential difference. What power rating is required? (c) A 100.0-\(\Omega\) and a 150.0-\(\Omega\) resistor, both rated at 2.00 W, are connected in series across a variable potential difference. What is the greatest this potential difference can be without overheating either resistor, and what is the rate of heat generated in each resistor under these conditions?

A circuit consists of a series combination of 6.00-k\(\Omega\) and 5.00-k\(\Omega\) resistors connected across a 50.0-V battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the 5.00-k\(\Omega\) resistor using a voltmeter having an internal resistance of 10.0 k\(\Omega\). (a) What potential difference does the voltmeter measure across the 5.00-k\(\Omega\) resistor? (b) What is the \(true\) potential difference across this resistor when the meter is not present? (c) By what percentage is the voltmeter reading in error from the true potential difference?

Two light bulbs have constant resistances of 400 \(\Omega\) and 800 \(\Omega\). If the two light bulbs are connected in series across a 120-V line, find (a) the current through each bulb; (b) the power dissipated in each bulb; (c) the total power dissipated in both bulbs. The two light bulbs are now connected in parallel across the 120-V line. Find (d) the current through each bulb; (e) the power dissipated in each bulb; (f) the total power dissipated in both bulbs. (g) In each situation, which of the two bulbs glows the brightest? (h) In which situation is there a greater total light output from both bulbs combined?

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 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance of each bulb does not vary with current. (\(Note:\) This description of a light bulb gives the power it dissipates when connected to the stated potential difference; that is, a 25-W, 120-V light bulb dissipates 25 W when connected to a 120-V line.) (a) Find the current through the bulbs. (b) Find the power dissipated in each bulb. (c) One bulb burns out very quickly. Which one? Why?

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