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Chapter 26: Problem 68

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?

Short Answer

Expert verified
The power dissipated when the three resistors are connected in parallel across the same potential difference is \(135.0 W\).

Step by step solution

01

Calculate total resistance for series circuit

Since the three resistors are identical and connected in series, total resistance \(R_s\) is simply three times the resistance of a single resistor. However, we do not know the exact value of the resistance of one resistor. Therefore, we denote it as \(R_s = 3R\).
02

Determine power for the series circuit

The power dissipated in the series circuit is given as \(45.0W\). From this, we construct the equation for power in terms of resistance as \(P_s = V^2/R_s\). Since we know \(P_s\), we can express voltage \(V\) as \(V = \sqrt{P_s \cdot R_s}\).
03

Calculate total resistance for parallel circuit

In a parallel circuit, the total resistance is the reciprocal of the sum of the reciprocals of each resistor's resistance. With three identical resistors, we can write the total resistance \(R_p\) as \(R_p = 1/(3/R)\).
04

Determine power for the parallel circuit

We then construct the power equation for the parallel circuit using the same voltage (as it doesn't change) and the parallel resistance we just calculated, \(P_p = V^2/R_p\). Substituting \(V = \sqrt{P_s \cdot R_s}\) and \(R_p = 1/(3/R)\), we can solve for \(P_p\).

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Key Concepts

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

Resistors in Series
When resistors are connected in series, it means they are linked one after another in a circuit. This type of connection is simple to understand because the current flows through each resistor sequentially. One primary characteristic of resistors in series is that the total resistance is simply the sum of all individual resistances.

For example, if you have three identical resistors with resistance \(R\), the total resistance \(R_s\) in series becomes \(3R\). This increased resistance affects the current passing through the circuit. Series circuits are beneficial when you need a higher resistance. However, they will also lead to a reduction in current flow due to the increased opposition to the electric current.
Resistors in Parallel
Parallel resistors represent an entirely different scenario compared to series resistors. In parallel configuration, all resistors are connected across the same two points, creating multiple pathways for the current to travel. This setup allows more current to pass through the circuit, unlike in series connection.

Calculating total resistance for resistors in parallel involves taking the reciprocal of the sum of the reciprocals of each resistor's resistance. In simple terms, for three identical resistors each having resistance \(R\), the total resistance \(R_p\) becomes \( R/3 \).

Parallel connections are advantageous for reducing the overall resistance which can be helpful if you want to maximize current flow in the circuit.
Power Dissipation
Power dissipation refers to the loss of electrical power typically in the form of heat within resistors as the electric current flows through them. It is an essential concept when analyzing circuits since every resistor will dissipate some power.

Power is calculated using the formula \(P = V^2/R\) where \(P\) is the power, \(V\) is the voltage, and \(R\) is the resistance. In both series and parallel circuits, understanding how to compute power dissipation is crucial for designing devices that can handle specific power levels without overheating or failing.
For the scenario of three resistors originally dissipating 45.0W in series, reconfiguring them in parallel will alter the power dissipated since the overall resistance in parallel decreases, increasing the total power dissipation through each resistor.
Ohm's Law
Ohm's Law forms the foundation for understanding electric circuits. This law states that the current through a conductor between two points is directly proportional to the voltage across the two points, and is inversely proportional to the resistance between them. Mathematically, this is written as \(V = IR\), where \(V\) is the voltage, \(I\) is the current, and \(R\) is the resistance.

Using Ohm's Law, you can analyze the relationships between voltage, current, and resistance in both series and parallel circuits. For instance, when you know any two values, you can calculate the third. Understanding how to apply Ohm's Law is crucial for solving real-world circuit problems because it helps you predict how changes in resistance or voltage will affect the current, and subsequently, the power dissipated in the circuit.

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Most popular questions from this chapter

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

The three-terminal device circled in Fig. \(\mathbf{P} 26.81\) is an NPN transistor. It works using three simple rules: I. The net current into the device is the same as the net current out of the device, so \(I_{\mathrm{C}}+I_{\mathrm{B}}=I_{\mathrm{E}} .\) II. The potential at point \(e\) is always \(0.60 \mathrm{~V}\) less than the potential at point \(b .\) Thus, \(V_{\text {in }}-V_{e}=0.60 \mathrm{~V}\). III. The current into point \(c\) is always a fixed multiple of the current into point \(b,\) so \(I_{\mathrm{C}}=\beta I_{\mathrm{B}},\) where \(\beta \gg 1\) is a parameter characteristic of the transistor. Use these rules to answer the questions. (a) What is the potential \(V_{\text {out }}\) in terms of \(V_{\text {in }},\) in the limit \(\beta \rightarrow \infty\) ? (Hint: Determine \(I_{E}\) in terms of \(V_{\text {in }}\), and use this to determine \(I_{\mathrm{C}}\) in terms of \(V_{\mathrm{in}}\) and \(\beta .\) The potential difference across the \(1.0 \mathrm{k} \Omega\) resistor can then be used to determine \(V_{\text {out }}\) Take the limit \(\beta \rightarrow \infty\) in this result. \((b)\) What value of \(V_{\text {in }}\) is needed so that \(V_{\text {out }}=7.5 \mathrm{~V} ?\) (c) If the input potential includes a small time-dependent "signal," so that \(V_{\text {in }}=15.0 \mathrm{~V}+v_{\text {in }}(t),\) then the output potential is \(V_{\text {out }}=7.5\) \(\mathrm{V}+G v_{\mathrm{in}}(t),\) where \(G\) is a "gain" factor. What is \(G\) for this circuit?

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?

The heating element of an electric dryer is rated at \(4.1 \mathrm{~kW}\) when connected to a \(240 \mathrm{~V}\) line. (a) What is the current in the heating element? Is 12 gauge wire large enough to supply this current? (b) What is the resistance of the dryer's heating element at its operating temperature? (c) At 11 cents per \(\mathrm{kWh}\), how much does it cost per hour to operate the dryer?

A capacitor is charged to a potential of \(12.0 \mathrm{~V}\) and is then connected to a voltmeter having a total internal resistance of \(3.40 \mathrm{M} \Omega\). After a time of 4.00 s the voltmeter reads \(3.0 \mathrm{~V}\). What are (a) the capacitance and (b) the time constant of the circuit?
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