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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

The most common material used for sandpaper, silicon carbide, is also widely used in electrical applications. One common device is a tubular resistor made of a special grade of silicon carbide called carborundum. A particular carborundum resistor (see the figure) consists of a thick-walled cylindrical shell (a pipe) of inner radius \(a=\) \(1.50 \mathrm{~cm},\) outer radius \(b=2.50 \mathrm{~cm},\) and length \(L=60.0 \mathrm{~cm} .\) The resistance of this carborundum resistor at \(20 .{ }^{\circ} \mathrm{C}\) is \(1.00 \Omega\). a) Calculate the resistivity of carborundum at room temperature. Compare this to the resistivities of the most commonly used conductors (copper, aluminum, and silver). b) Carborundum has a high temperature coefficient of resistivity: \(\alpha=2.14 \cdot 10^{-3} \mathrm{~K}^{-1} .\) If, in a particular application, the carborundum resistor heats up to \(300 .{ }^{\circ} \mathrm{C},\) what is the percentage change in its resistance between room temperature \(\left(20 .{ }^{\circ} \mathrm{C}\right)\) and this operating temperature?

What is (a) the conductance and (b) the radius of a \(3.5-\mathrm{m}\) -long iron heating element for a \(110-\mathrm{V}, 1500-\mathrm{W}\) heater?

If the current through a resistor is increased by a factor of \(2,\) how does this affect the power that is dissipated? a) It decreases by a factor of 4 . b) It increases by a factor of 2 . c) It decreases by a factor of 8 . d) It increases by a factor of 4 .

In an emergency, you need to run a radio that uses \(30.0 \mathrm{~W}\) of power when attached to a \(10.0-\mathrm{V}\) power supply. The only power supply you have access to provides \(25.0 \mathrm{kV}\) but you do have a large number of \(25.0-\Omega\) resistors. If you want the power to the radio to be as close as possible to \(30.0 \mathrm{~W}\), how many resistors should you use, and how should they be connected (in series or in parallel)?

A battery has a potential difference of \(14.50 \mathrm{~V}\) when it is not connected in a circuit. When a \(17.91-\Omega\) resistor is connected across the battery, the potential difference of the battery drops to \(12.68 \mathrm{~V}\). What is the internal resistance of the battery?

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