# chap25 - ELECTRIC CIRCUITS 25 EXERCISES Section 25.1...

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25.1 ELECTRIC CIRCUITS 25 EXERCISES 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. INTERPRET This problem is about how various circuit elements can be connected to form a closed series circuit. DEVELOP In a series circuit, the same current must flow through all elements. EVALUATE One possibility is shown below. The order of elements and the polarity of the battery connections are not specified. ASSESS 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 and in series; one with just and one with the battery (an ideal emf in series with the internal resistance). 1 R 2 R 3 ; R 17. INTERPRET This problem explores the connection between the emf of a battery and the energy it delivers. DEVELOP Electromotive force, or emf, is defined as work per unit charge, /. Wq ε = EVALUATE Substituting the values given in the problem statement, we find the emf to be 27 J 9 V 3 C W q == =

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25.2 Chapter 25 ASSESS 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, , PI ε = therefore 4.5 kJ, It = or 3 4.5 kJ/(1.5 V)(0.60 A) 5 10 s 1.39 h. t == × = 19. INTERPRET This problem is about the chemical energy used up in the battery for the work done. DEVELOP The power delivered by an emf is . = Therefore, if the voltage and current remain constant, then the energy converted would be . WP tI t EVALUATE Substituting the values given, the energy used in (5 A)(12 V)(3600 s) 216 kJ WI t = ASSESS 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 21. INTERPRET This problem is about connecting two resistors in parallel. DEVELOP The equivalent resistance of two resistors connected in parallel can be found by Equation 25.3a: parallel 1 2 11 RR =+ 1 R The equation allows us to determine when 2 R parallel R and are known. 1 R EVALUATE The solution for R 2 in Equation 25.3a is 1 parallel 2 1 parallel (56 k )(45 k ) 229 k 56 k 45 k R ΩΩ = −Ω Ω Ω ASSESS Our result shows that This is consistent with the fact that the equivalent resistance 2 parallel . > parallel R is smaller than and 1 R 2 . R 22. The starter circuit contains all the resistances in series, as in Figure 25.10. (We assume L R includes the resistance of the cables, connections, etc., as well as that of the motor.) With the defective starter, int 6 V
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## This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.

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chap25 - ELECTRIC CIRCUITS 25 EXERCISES Section 25.1...

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