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Chapter 21: Q4CQ (page 771)

Why is the power dissipated by a closed switch, such as in Figure 21.43, small?

Short Answer

Expert verified

Due to the low resistance of a closed switch, the power dissipated is little.

Step by step solution

01

Definition of voltage

Voltage: It's the difference between two points’ electric potentials.

02

Given

Resistance of the switch when closedRclosed=0Ω

Resistance of switch when openRopen=∞Ω

Emf of the battery isE

The internal resistance of the battery isr

External resistance connected in series with the battery isR

Current in the circuit when the switch is closedlclosed

Current in the circuit when the switch is openlopen

03

Finding why the power dissipated by the closed switch is small

The power dissipated is given :

P=lclosed2RclosedP=lclosed20ΩP=0W

Resistance is directly proportional to power. As a result, the lower the resistance, the lower the power dissipated.

Therefore, the resistance of a closed switch is so low, that the power dissipated is little.

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Most popular questions from this chapter

The \({\rm{RC}}\) time constant in heart defibrillation is crucial to limiting the time the current flows. If the capacitance in the defibrillation unit is fixed, how would you manipulate resistance in the circuit to adjust the \({\rm{RC}}\) constant \(\tau \)? Would an adjustment of the applied voltage also beaded to ensure that the current delivered has an appropriate value?

A heart pacemaker fires 72 times a minute, each time a 25.0-nF capacitor is charged (by a battery in series with a resistor) to 0.632 of its full voltage. What is the value of the resistance?

Apply the junction rule at point a in Figure 21.52.

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A heart defibrillator being used on a patient has an RC time constant of 10.0msdue to the resistance of the patient and the capacitance of the defibrillator. (a) If the defibrillator has an8.00-μFcapacitance, what is the resistance of the path through the patient? (You may neglect the capacitance of the patient and the resistance of the defibrillator.) (b) If the initial voltage is12.0KV, how long does it take to decline to6.00×102V?
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