chapter18 - 142 Chapter 18 PROBLEM SOLUTIONS 18.1 From V =...

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142 Chapter 18 PROBLEM SOLUTIONS 18.1 From Δ VIRr =+ ( ) , the internal resistance is r V I R =− = = Δ ΩΩ 900 72 0 4 92 . .. V 0.117 A 18.2 (a) When the three resistors are in series, the equivalent resistance of the circuit is RR R R eq =++= ( ) = 123 390 27 . (b) The terminal potential difference of the battery is applied across the series combination of the three 90 . Ω resistors, so the current supplied by the battery and the current through each resistor in the series combination is I V R == = Δ Ω eq V 27 A 12 044 . (c) If the three 90 Ω are now connected in parallel with each other, the equivalent resistance is 11 90 1 1 3 R eq .... ΩΩΩ Ω or R eq 3 30 . . Ω Ω When this parallel combination is connected to the battery, the potential difference across each resistor in the combination is Δ V = 12 V, so the current through each of the resistors is I V R = Δ Ω 12 13 V 9.0 A . 18.3 For the bulb in use as intended, R V bulb V W = ( ) = ( ) = Δ Ω 22 120 75 0 192 P . Now, assuming the bulb resistance is unchanged, the current in the circuit shown is I V R ++ = Δ Ω eq V 0.800 A 120 192 0 800 0 620 . . and the actual power dissipated in the bulb is P ( ) ( ) = IR 2 2 0 620 192 73 8 bulb A W Ω 18.4 (a) When the 800 .- Ω resistor is connected across the 9.00-V terminal potential difference of the battery, the current through both the resistor and the battery is I V R = Δ Ω 113 . . V 8.00 A (b) The relation between the emf and the terminal potential difference of a battery supplying current I is Δ VI r ε , where r is the internal resistance of the battery. Thus, if the battery has r = 015 Ω and maintains a terminal potential difference of Δ V = V while sup- plying the current found above, the emf of this battery must be = + ( )( ) ( ) ΔΩ r 900 017 . . . V A = 917 .
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Direct-Current Circuits 143 18.5 (a) The equivalent resistance of the two parallel resistors is R p =+ = 1 700 1 10 0 412 1 .. . ΩΩ Ω Thus, RR R R ab p =++= + + ( ) = 49 400 412 900 171 ... . (b) I V R ab ab ab = ( ) == Δ Ω 34 0 17 1 199 . . . V A, so II . A Also, ΔΩ VI R p ab p ( ) ( )( ) = 818 . A V Then, I V R p 7 7 117 = ( ) Δ Ω . . V 7.00 A and I V R p 10 10 0 818 = ( ) Δ Ω . . V 10.0 A 18.6 (a) The parallel combination of the 60 and 12 resistors has an equivalent resistance of 11 60 1 12 21 12 1 R p = + ΩΩ Ω or R p 1 12 3 40 Ω Ω . Similarly, the equivalent resistance of the 40 and 8.0 parallel combination is 1 80 2 R p =+= + or R p 2 8 3 = Ω The total resistance of the series combination between points a and b is then R ab p p + = + + = 12 50 3 35 3 . . Ω Ω Ω (b) If Δ V ab = 35 V, the total current from a to b is IV R ab ab ab ( ) = 35 3 0 V 35 3 A . and the potential differences across the two parallel combinations are R pa b p 30 ( )( ) = A V and R b p 22 3 ( ) = . . . A V so the individual currents through the various resistors are: p 12 1 12 1 0 A . ; p 61 20 A ; ab 5 . A ; p 82 10 A ; and p 42 A
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144 Chapter 18 18.7 The equivalent resistance of the parallel combination of three identical resistors is 11113 123 RR RRR p =++= or R R p = 3 The total resistance of the series combination between points a and b is then R R R ab p =+ += + = 2 3 7 3 18.8 (a) The equivalent resistance of this F rst parallel combination is 11 10 0 1 500 1 R p =+ .. ΩΩ or R p 1 333 = .
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This note was uploaded on 04/23/2010 for the course DS SAD taught by Professor Asd during the Spring '10 term at A.T. Still University.

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chapter18 - 142 Chapter 18 PROBLEM SOLUTIONS 18.1 From V =...

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