Physics Solution Manual for 1100 and 2101

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Unformatted text preview: 2 t = cm(T − 0.0 °C ) + mLf t 1078 ELECTRIC CIRCUITS The power produced by an electric heater is, according to Equation 20.6a, P = IV. Substituting this expression for P into the equation above and solving for the current I, we get cm(T − 0.0 °C ) + mLf I= tV (4186 J/kg ⋅ C°)(660 kg)(10.0 C°) + (660 kg)(33.5 × 104 J/kg) = 32 A 3600 s (9.0 h) (240 V) h ______________________________________________________________________________ I= 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 0 R0 . When the additional resistor R is inserted in series, assuming that the battery remains the same, the voltage across the new combination is given by V = I ( R + R0 ) . Since V is the same in both cases, we can write I 0 R0 = I ( R + R0 ) . This expression can be solved for R0 . SOLUTION Solving for R0 , we have I 0 R0 – IR0 = IR R0 ( I 0 – I ) = IR or Therefore, we find that IR (12.0 A)(8.00 Ω) = = 32 Ω I0 – I 15.0 A–12.0 A ______________________________________________________________________________ R0 = 42. REASONING According to Equation 20.2, the resistance R of the resistor is equal to the voltage VR across it divided by the current I, or R = VR /I . Since the resistor, the lamp, and the voltage source are in series, the voltage across the resistor is VR = 120.0 V − VL, where VL is the voltage across the lamp. Thus, the resistance is R= 120.0 V − VL I Since VL is known, we need only determine the current in the circuit. Since we know the voltage VL across the lamp and the power P dissipated by it, we can use Equation 20.6a to find the current: I = P / VL . The resistance can be written as Chapter 20 Problems R= 1079 120.0 V − VL P VL SOLUTION Substituting the known values for VL and P into the equation above, the resistance is 120.0 V − 25 V = 4.0 × 101 Ω R= 60.0 W 25 V ______________________________________________________________________________ 43. SSM REASONING The equivalent series resistance Rs is the sum of the resistances of the three resistors. The potential difference V can be determined from Ohm's law according to V = IRs . SOLUTION a. The equivalent resistance is Rs = 25 Ω + 45 Ω + 75 Ω = 145 Ω b. The potential difference across the three resistors is V = IRs = (0.51 A)(145 Ω) = 74 V ______________________________________________________________________________ 44. REASONING Ohm’s law relates the resistance R of either resistor to the current I in it and the voltage V across it: R= V I (20.2) Because the two resistors are in series, they must have the same current I. We will, therefore, apply Equation 20.2 to the 86-Ω resistor to determine the current I. Following that, we will use Equation 20.2 again, to obtain the potential difference across the 67-Ω resistor. SOLUTION Let R1 = 86 Ω be the resistance of the first resistor, which has a potential difference of V1 = 27 V across it. The current I in this resistor, from Equation 20.2, is I= V1 R1 (1) 1080 ELECTRIC CIRCUITS Let R2 = 67 Ω be the resistance of the second resistor. Again employing Equation 20.2, the potential difference V2 across this resistor is given by V2 = IR2 (2) Since the current in both resistors is the same, substituting Equation (1) into Equation (2) yields V ( 27 V ) 67 Ω = 21 V V2 = IR2 = 1 R2 = ( ) R ( 86 Ω ) 1 ______________________________________________________________________________ 45. REASONING Since the two resistors are connected in series, they are equivalent to a single equivalent resistance that is the sum of the two resistances, according to Equation 20.16. Ohm’s law (Equation 20.2) can be applied with this equivalent resistance to give the battery voltage. SOLUTION According to Ohm’s law, we find V = IRs = I ( R1 + R2 ) = ( 0.12 A )( 47 Ω + 28 Ω ) = 9.0 V ______________________________________________________________________________ 46. REASONING The circuit containing the light bulb and resistor is shown in the drawing. The resistance R1 of the light bulb is related to the power delivered to it by R1 = P / I 2 1 (Equation 20.6b), where I is the current in the circuit. The power is known, and the current can be obtained from Ohm’s law as the voltage V of the source divided by the equivalent resistance RS of the series circuit: I = V / RS . Since the two resistors are wired in series, the equivalent resistance is the sum of the resistances, or RS = R1 + R2 . Light bulb R2 = 144 R1 P1 = 23.4 W 120.0 V SOLUTION The resistance of the light bulb is R1 = P 1 I2 (20.6b) Chapter 20 Problems 1081 Substituting I = V / RS (Equation 20.2) into Equation 20.6b gives R1 = P 1 I2 = P 1 (V / RS )2 = 2 P RS 1 (1) V2 The equivalent resistance of the two resistors wired in series is RS = R1 + R2 (Equation 20.16). Substituting this expression for RS into Equation (1) yields R1 = 2 P RS 1 V2 = P ( R1 + R2 ) 1 2 V2 Algebraically rearranging this equation, we find that V2 2 2 R1 + 2 R2 − R + R2 =...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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