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Chapter 26: Problem 3

A resistor with \(R_{1}=25.0 \Omega\) is connected to a battery that has negligible internal resistance and electrical energy is dissipated by \(R_{1}\) at a rate of \(36.0 \mathrm{~W}\). If a second resistor with \(R_{2}=15.0 \Omega\) is connected in series with \(R_{1},\) what is the total rate at which electrical energy is dissipated by the two resistors?

Short Answer

Expert verified
The total rate at which electrical energy is dissipated by the two resistors is 57.6W.

Step by step solution

01

Find the current

First, we need to find the current. Given that the power dissipated by the first resistor \(P_{1}\) is 36.0W and its resistance \(R_{1}\) is 25.0惟, the current \(I\) can be found by rearranging the formula \(P_{1} = I^{2} * R_{1}\) to \(I = \sqrt { \frac {P_{1}} {R_{1}} }\), which gives us \(I = \sqrt { \frac {36.0W} {25.0 \Omega} }\) = 1.2A.
02

Find the total resistance

Next, add the values of \(R_{1}\) and \(R_{2}\) together to get the total resistance \(R_{T} = R_{1} + R_{2} = 25.0 \Omega + 15.0 \Omega = 40.0 \Omega.\)
03

Find the total power dissipated

Finally, using the current from Step 1 and the total resistance from Step 2, apply the power formula again \(P_{T} = I^{2} * R_{T}\) to get \(P_{T} = (1.2A)^{2} * 40.0 \Omega = 57.6W.\)

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Key Concepts

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

series circuit
When resistors are connected in a series circuit, the electrical current that flows through them is the same across each component.
This is because there is only one path for the current to take.
But the voltage across each resistor may be different.
  • In a series circuit, the total resistance is the sum of the resistances of each resistor.
  • This means you'll have more resistance compared to a single resistor.
  • Series circuits are commonly used when you want the same current to flow through all components.
For example, in the exercise where two resistors are connected in series, both resistors have the same current of 1.2 A flowing through them. The series connection ensures that the total resistance is simply the sum of the individual resistances: 25 惟 for the first resistor and 15 惟 for the second resistor, resulting in a total resistance of 40 惟.
resistor power dissipation
Resistor power dissipation is the process through which resistors convert electrical energy into heat energy. Power dissipation in a resistor is an important concept for understanding how resistors work and how much heat they generate.Power dissipation can be calculated using the formula:\[ P = I^2 * R \]where:
  • \( P \) is the power dissipated by the resistor
  • \( I \) is the current flowing through the resistor
  • \( R \) is the resistance of the resistor
The exercise gives us an example of power dissipation with a figure of 36W for a single resistor with 25 惟. With a current of 1.2 A, we later calculate that the two resistors together dissipate 57.6 W when in series.
When designing circuits, it's crucial to ensure that resistors can handle the amount of power dissipated to avoid overheating.
total resistance calculation
Calculating the total resistance in series circuits is straightforward and involves adding up the resistances of each resistor.
  • This is because the total resistance in a series circuit is the sum total of all individual resistances.
  • It reflects the cumulative opposition to the flow of current through the circuit.
In the given problem, for instance, we have two resistors: one with 25 惟 and another with 15 惟. The total resistance is calculated as:\[ R_T = R_1 + R_2 = 25 \, \Omega + 15 \, \Omega = 40 \, \Omega \]Understanding this concept allows you to easily predict how the circuit behavior changes when you add more resistors in series, as it increases the total resistance, reducing the current if the voltage remains constant.

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Most popular questions from this chapter

Two light bulbs have constant resistances of \(400 \Omega\) and \(800 \Omega .\) If the two light bulbs are connected in series across a \(120 \mathrm{~V}\) 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 \(120 \mathrm{~V}\) 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?

Two capacitor plates with area \(A\) are separated by a distance \(d\). The space between the plates is filled with dielectric material with dielectric constant \(K\) and resistivity \(\rho\). This capacitor is attached to a battery that supplies a constant potential \(\mathcal{E},\) as shown in Fig. \(\mathbf{P} 26.69,\) and is fully charged. At time \(t=0\) the switch \(S\) is opened. Owing to the nonzero value of \(\rho,\) this capacitor discharges by leaking. We can model this device as a capacitor and a resistor in parallel. (a) Analyze this circuit and determine the time constant, characterizing the discharge in terms of the parameters given above. Now assume the capacitor has plates with area \(A=1.00 \mathrm{~cm}^{2}\) separated by \(d=30.0 \mu \mathrm{m}\) and is filled with a ceramic of dielectric constant \(K=8.70\) and resistivity \(\rho=3.10 \times 10^{12} \Omega \cdot \mathrm{m}\) (b) If \(\mathcal{E}=5.00 \mathrm{~V},\) what is the charge on the capacitor at \(t=0 ?\) (c) At what time will the capacitor have half of its original charge? (d) What is the magnitude of the leaking current at that point?

To measure the capacitance \(C\) of a capacitor, you attach the capacitor to a battery and wait until it is fully charged. You then disconnect the capacitor from the battery and let it discharge through a resistor of resistance \(R\). You measure the time \(T_{1 / 2}\) that it takes the voltage across the resistor to decrease to half its initial value at the instant that the connection to the capacitor is first completed. You repeat this for several different resistors. You plot the data as \(T_{1 / 2}\) versus \(R\) and find that they lie close to a straight line that has slope \(5.00 \mu \mathrm{F}\). What is the capacitance \(C\) of the capacitor?

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 \(15 \mathrm{k} \Omega\) resistor is \(5.0 \mathrm{~W},\) what is the maximum allowable potential difference across the terminals of the resistor? (b) \(A\) \(9.0 \mathrm{k} \Omega\) resistor is to be connected across a \(120 \mathrm{~V}\) potential difference. What power rating is required? (c) \(\mathrm{A} 100.0 \Omega\) and a \(150.0 \Omega\) resistor, both rated at \(2.00 \mathrm{~W}\), 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?

Three identical resistors are connected in series. When a certain potential difference is applied across the combination, the total power dissipated is \(45.0 \mathrm{~W}\). What power would be dissipated if the three resistors were connected in parallel across the same potential difference?
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