25_InstSolManual_PC - ELECTRIC CIRCUITS E XERCISES Section...

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Unformatted text preview: ELECTRIC CIRCUITS E XERCISES Section 25.1 Circuits, Symbols, and Electromotive Force 14. A literal reading of the circuit specifications results in connections like those in sketch (a) . Because the connecting wires are assumed to have no resistance (a real wire is represented by a separate resistor), a topologically equivalent circuit diagram is shown in sketch (b) . 15. I NTERPRET This problem is about how various circuit elements can be connected to form a closed series circuit. D EVELOP In a series circuit, the same current must flow through all elements. E VALUATE One possibility is shown below. The order of elements and the polarity of the battery connections are not specified. A SSESS An important feature about a series circuit is that the current through all the components must be the same. With two batteries, the direction of the current flow is determined by the polarity of the larger of the two voltage ratings. 16. The circuit has three parallel branches: one with 1 R and 2 R in series; one with just 3 ; R and one with the battery (an ideal emf in series with the internal resistance). 17. I NTERPRET This problem explores the connection between the emf of a battery and the energy it delivers. D EVELOP Electromotive force, or emf, is defined as work per unit charge, / . W q ε = E VALUATE Substituting the values given in the problem statement, we find the emf to be 27 J 9 V 3 C W q ε = = = 25.1 25 25.2 Chapter 25 A SSESS For an ideal battery with zero internal resistance, the emf is equal to the terminal voltage (potential difference across the battery terminals). 18. The average power, supplied by the battery to the bulb, multiplied by the time equals the energy capacity of the battery. For an ideal battery, , P I ε = therefore 4.5 kJ, It ε = or 3 4.5 kJ/(1.5 V)(0.60 A) 5 10 s 1.39 h. t = = × = 19. I NTERPRET This problem is about the chemical energy used up in the battery for the work done. D EVELOP The power delivered by an emf is . P I ε = Therefore, if the voltage and current remain constant, then the energy converted would be . W Pt I t ε = = E VALUATE Substituting the values given, the energy used in (5 A)(12 V)(3600 s) 216 kJ W I t ε = = = A SSESS The result makes sense; the energy used up is proportional to the current drawn, the emf, and the duration the headlights were left on. Section 25.2 Series and Parallel Circuits 20. From Equations 25.1 and 25.3b, 22 k 47(39 k )/(47 39) 43.3 k . R = Ω + Ω + = Ω 21. I NTERPRET This problem is about connecting two resistors in parallel. D EVELOP The equivalent resistance of two resistors connected in parallel can be found by Equation 25.3a: parallel 1 2 1 1 1 R R R = + The equation allows us to determine 2 R when parallel R and 1 R are known....
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This note was uploaded on 09/12/2009 for the course PH PH101 taught by Professor Prof.kim during the Spring '09 term at Korea Advanced Institute of Science and Technology.

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25_InstSolManual_PC - ELECTRIC CIRCUITS E XERCISES Section...

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