Electric Circuits

Discussion Questions

1: A current of 12 A flows for 2.5 minutes to charge a battery. How much charge is transferred to the battery in this time?

Solution:

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2: The resistivity of copper is 1.72 × 10-8 Ω.m. Determine the resistance of a copper wire that is 1.3 m long and has a diameter of 2.1 mm.

Solution:

Chapter 20 DiscussionJanuary-03-15 8:58 PM

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3: A potential difference of 27 V is applied to the wire in Q2. Determine the current that flows in the wire.

Solution:

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4: If a copper wire is replaced by another copper wire that has twice the length and twice the diameter, how does the resistance change?

Solution:

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5: The voltage drop across the terminals of a light bulb is 120 V, and the light bulb dissipates 60 W of power. (a) Determine the current flowing through the bulb.(b) Determine the resistance of the bulb.(c) Determine the amount of energy dissipated by the bulb in 24 h.(d) Determine how much charge flows through the bulb in 24 h.(e) Determine the cost to run the light bulb for 24 h if electricity costs 0.10 dollars per kW.h.

Solution:

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6: A 20-Ω resistor is connected across the terminals of a 10-V battery. Some current flows through the resistor, and some power is dissipated in the resistor. Then the resistor is replaced by a 40-Ωresistor. (a) Who is correct? Explain.Student 1: The voltage drop across the 40-Ω resistor is the same as the voltage drop across the 20-Ω resistor.Student 2: The voltage drop across the 40-Ω resistor is twice as much as the voltage drop across the 20-Ω resistor. Ch20DClass Page 4

as the voltage drop across the 20-Ω resistor.Student 3: The voltage drop across the 40-Ω resistor half as much as the voltage drop across the 20-Ω resistor.(b) Who is correct? Explain.Student 1: The current through the 40-Ω resistor is the same as the current through the 20-Ω resistor.Student 2: The current through the 40-Ω resistor is twice as much as the current through the 20-Ω resistor.Student 3: The current through the 40-Ω resistor half as much as the current through the 20-Ω resistor.(c) Who is correct? Explain.Student 1: The power dissipated in the 40-Ω resistor is the same as the power dissipated in the 20-Ω resistor.Student 2: The power dissipated in the 40-Ω resistor is twice as much as the power dissipated in the 20-Ω resistor.Student 3: The power dissipated in the 40-Ω resistor half as much as the power dissipated in the 20-Ω resistor.

Solution:

7: A battery is rated at 200 A.h. (a) Determine the total charge that the battery can provide.(b) Determine the maximum current that the battery could provide for 48 minutes.

Solution: (a) The A.h is a unit of charge, so we can conclude that the battery can deliver 200 A.h of charge.

If you wish to convert this to coulombs, then:

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Series circuits

Example: Determine the current through the resistor and the power dissipated in the resistor.

Solution:

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Example: Determine the current flowing through each resistor and the power dissipated in each resistor.

Solution:

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Example: Determine the equivalent resistance when resistors of 5 Ω, 8 Ω, and 11 Ω are connected in series.

Solution:

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Example: Determine the current through each resistor and the power dissipated by each resistor.

Solution:

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Example: Determine the current through each resistor and the power dissipated by each resistor.

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Solution: This example is similar to the previous one. The only difference is that one of the batteries has been connected with the opposite orientation.

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Parallel circuits

Example: Determine the current through each resistor and the power dissipated in each resistor. Compare to an earlier example that was similar but the resistors were connected in series.

Solution:

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Example: Determine the current through each resistor and the power dissipated in each resistor.

Solution:

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Example: Determine the effective resistance for each set of resistors connected in parallel.a: 10 Ω, 10 Ωb: 10 Ω, 10 Ω, 10 Ωc: 10 Ω, 10 Ω, 10 Ω, 10 Ωd: 10 Ω, 10 Ω, 10 Ω, 10 Ω, 10 Ωe: 2 Ω, 3 Ω, 6 Ω

Solution:

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Example: Determine the current through each resistor and the power dissipated in each resistor.

Solution: Note that if the 1 Ω resistor were not present, the current in the 100 Ω resistor would be 100/100 = 1 A. Now notice how dramatically the current changes when the 1 Ω resistor is included in the circuit:

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More complex circuits (i.e., combinations of series and parallel connections)

1: Practice determining effective resistance

2: solving complex circuits … examples

3: challenges (resistor cube, tetrahedron net, infinite network)

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