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Chapter 25: Problem 17

Show that for resistors connected in series, it is always the highest resistance that dissipates the most power, while for resistors connected in parallel, it is always the lowest resistance that dissipates the most power.

Short Answer

Expert verified
Question: Explain which resistor dissipates the most power in a series and parallel circuit, and justify your answer.

Step by step solution

01

Apply Ohm's Law for series resistors

The total resistance in a series circuit is obtained by simply adding all the individual resistances. Let's assume we have n resistors connected in series with resistances R1, R2, ..., Rn. The total resistance R_total = R1 + R2 + ... + Rn.
02

Calculate total current in the circuit

Using Ohm's Law (V = I * R), we can calculate the total current in the circuit. If V is the voltage across the circuit, the total current I = V / R_total.
03

Calculate power dissipation for each resistor

Power dissipation for a resistor can be calculated using P = I^2 * R. Since the current flowing through each resistor in a series circuit is the same, the resistor with the highest resistance will dissipate the most power. **For resistors connected in parallel:**
04

Apply Ohm's Law for parallel resistors

In a parallel circuit, the inverse of the total resistance is equal to the sum of the inverses of the individual resistances. Let's assume we have n resistors connected in parallel with resistances R1, R2, ..., Rn. The total resistance R_total = 1 / (1/R1 + 1/R2 + ... + 1/Rn).
05

Calculate voltage across each resistor

For parallel connected resistors, the voltage across each resistor is the same as the total applied voltage. If V is the voltage across the circuit, the voltage across each resistor is also V.
06

Calculate the current through each resistor

Using Ohm's Law (V = I * R), we can find the current through each resistor. For the ith resistor, the current I_i = V / R_i.
07

Calculate power dissipation for each resistor

Power dissipation for a resistor can be calculated using P = I^2 * R. Since the voltage across each resistor in a parallel circuit is the same, the resistor with the lowest resistance will have the highest current and thus, dissipate the most power. By following these steps, we have shown that in a series circuit, the resistor with the highest resistance dissipates the most power, and in a parallel circuit, the resistor with the lowest resistance dissipates the most power.

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Most popular questions from this chapter

Show that the power supplied to the circuit in the figure by the battery with internal resistance is maximum when the resistance of the resistor in the circuit, \(R\), is equal to \(R_{i}\). Determine the power supplied to \(R\). For practice, calculate the power dissipated by a \(12.0-\mathrm{V}\) battery with an internal resistance of \(2.00 \Omega\) when \(R=1.00 \Omega, R=2.00 \Omega,\) and \(R=3.00 \Omega\)

A light bulb is connected to a source of emf. There is a \(6.20 \mathrm{~V}\) drop across the light bulb, and a current of 4.1 A flowing through the light bulb. a) What is the resistance of the light bulb? b) A second light bulb, identical to the first, is connected in series with the first bulb. The potential drop across the bulbs is now \(6.29 \mathrm{~V},\) and the current through the bulbs is \(2.9 \mathrm{~A}\). Calculate the resistance of each light bulb. c) Why are your answers to parts (a) and (b) not the same?

When a \(40.0-V\) emf device is placed across two resistors in series, a current of \(10.0 \mathrm{~A}\) is flowing in each of the resistors. When the same emf device is placed across the same two resistors in parallel, the current through the emf device is \(50.0 \mathrm{~A}\). What is the magnitude of the larger of the two resistances?

Why do light bulbs typically burn out just as they are turned on rather than while they are lit?

A potential difference of \(V=0.500 \mathrm{~V}\) is applied across a block of silicon with resistivity \(8.70 \cdot 10^{-4} \Omega \mathrm{m}\). As indicated in the figure, the dimensions of the silicon block are width \(a=2.00 \mathrm{~mm}\) and length \(L=15.0 \mathrm{~cm} .\) The resistance of the silicon block is \(50.0 \Omega\), and the density of charge carriers is \(1.23 \cdot 10^{23} \mathrm{~m}^{-3}\) Assume that the current density in the block is uniform and that current flows in silicon according to Ohm's Law. The total length of 0.500 -mm-diameter copper wire in the circuit is \(75.0 \mathrm{~cm},\) and the resistivity of copper is \(1.69 \cdot 10^{-8} \Omega \mathrm{m}\) a) What is the resistance, \(R_{w}\) of the copper wire? b) What are the direction and the magnitude of the electric current, \(i\), in the block? c) What is the thickness, \(b\), of the block? d) On average, how long does it take an electron to pass from one end of the block to the other? \(?\) e) How much power, \(P\), is dissipated by the block? f) In what form of energy does this dissipated power appear?
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