# HW5 - 1 ELECTRIC CIRCUITS HW5 Solutions 23 REASONING AND...

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Unformatted text preview: 1 ELECTRIC CIRCUITS HW5 Solutions 23. REASONING AND SOLUTION a. According to Equation 20.6c, the resistance is ( 29 2 2 12 V 4.4 33 W V R P = = = Ω b. According to Equation 20.6a, the current is 33 W 2.8 A 12 V P I V = = = ______________________________________________________________________________ 25. REASONING The total cost of keeping all the TVs turned on is equal to the number of TVs times the cost to keep each one on. The cost for one TV is equal to the energy it consumes times the cost per unit of energy (\$0.12 per kW ⋅ h). The energy that a single set uses is, according to Equation 6.10b, the power it consumes times the time of use. SOLUTION The total cost is ( 29 ( 29 ( 29 ( 29 Total cost = 110 million sets Cost per set \$0.12 110 million sets Energy in kW h used per set 1 kW h = ⋅ ⋅ The energy (in kW ⋅ h) used per set is the product of the power and the time, where the power is expressed in kilowatts and the time is in hours: ( 29 ( 29 1 kW Energy used per set = 75 W 6.0 h 1000 W Pt = (6.10b) The total cost of operating the TV sets is ( 29 ( 29 ( 29 6 1 kW \$0.12 Total cost = 110 million sets 75 W 6.0 h \$5.9 10 1000 W 1 kW h = × ⋅ ______________________________________________________________________________ 41. SSM REASONING Using Ohm's law (Equation 20.2) we can write an expression for the voltage across the original circuit as V I R = . When the additional resistor R is inserted in series, assuming that the battery remains the same, the voltage across the new combination is 2 ELECTRIC CIRCUITS given by ( ) V I R R = + . Since V is the same in both cases, we can write ( ) I R I R R = + . This expression can be solved for R . SOLUTION Solving for R , we have – or ( – ) I R IR IR R I I IR = = Therefore, we find that (12.0 A)(8.00 ) 32 – 15.0 A–12.0 A IR R I I Ω = = = Ω ______________________________________________________________________________ 43. SSM REASONING The equivalent series resistance R s is the sum of the resistances of the three resistors. The potential difference V can be determined from Ohm's law according to V = IR s . SOLUTION a. The equivalent resistance is R s = 25 Ω + 45 Ω + 75 Ω = 145 Ω b. The potential difference across the three resistors is V = IR s = (0.51 A)(145 Ω ) = 74 V ______________________________________________________________________________ 51. SSM REASONING AND SOLUTION Since the circuit elements are in parallel, the equivalent resistance can be obtained directly from Equation 20.17: 1 R p = 1 R 1 + 1 R 2 = 1 16 Ω + 1 8.0 Ω or R p = 5.3 Ω ______________________________________________________________________________ 65. SSM REASONING When two or more resistors are in series, the equivalent resistance is given by Equation 20.16: R s = R 1 + R 2 + R 3 + . . . . Likewise, when resistors are in parallel, the expression to be solved to find the equivalent resistance is given by Equation...
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HW5 - 1 ELECTRIC CIRCUITS HW5 Solutions 23 REASONING AND...

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