SOLUTION The energy used by the dryer is. 60 s 1.00 min = J. Energy = Pt = IVt = (16 A)(240 V)(45 min) For the computer, we have

Transcription

1 3. SSM REASONING According to Equation 6.1b, the energy used is Energy = Pt, where P is the power and t is the time. According to Equation.6a, the power is P = IV, wherei is the current and V is the voltage. Thus, Energy = IVt, and we apply this result first to the dryer and then to the computer. SOLUTION The energy used by the dryer is Energy = Pt = IVt = (16 A)(4 V)(45 min) For the computer, we have 6 s 1. min Converts minutes to seconds = J Energy = J = IVt = (.7 A) ( 1 V)t Solving for t we find ( ) ( )( ) J 1. h t = = s = s = 8.9 h.7 A 1 V 36 s

2 7. SSM REASONING AND SOLUTION As a function of temperature, the resistance of the wire is given by Equation.5: R = R 1 α( T T ) +,whereα is the temperature coefficient of resistivity. From Equation.6c, we have P = V / R. Combining these two equations, we have V P P = = R 1 α T T α + + ( ) 1 ( T T ) where P = V /R, since the voltage is constant. But P = 1 P,sowefind Solving for T,wefind P P = or = 1+ α 1 + α ( T T ) ( T T ) T 1 1 = + T = + 8 = 5 C 1 α.45 (C )

3 39. SSM REASONING Using Ohm's law (Equation.) we can write an expression for the voltage across the original circuit as V = IR. 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+ R ). Since V is the same in both cases, we can write IR = IR ( + R). This expression can be solved for R. SOLUTION Solving for R,wehave Therefore, we find that I R IR = IR or R ( I I) = IR IR (1. A)(8. Ω) R = = = 3 Ω I I 15. A 1. A

4 41. REASONING AND SOLUTION The equivalent resistance of the circuit is R s = R 1 + R = 36. Ω Ω = 54. Ω Ohm's law for the circuit gives I = V/R s = (15. V)/(54. Ω) =.78 A a. Ohm's law for R 1 gives V 1 = (.78 A)(36. Ω) = 1. V b. Ohm's law for R gives V = (.78 A)(18. Ω) = 5. V

5 44. REASONING AND SOLUTION Each resistor can tolerate a current of no more than I P.5 W = = =.73 A R 47 Ω Ohm's law applied to a series circuit containing N such resistors gives V = IR s = INR,so Only three resistors can be used. N = V IR = 9. V (.73 A) 47 Ω ( ) =.6

6 53. SSM REASONING Since the resistors are connected in parallel, the voltage across each one is the same and can be calculated from Ohm's Law (Equation.: V = IR). Once the voltage across each resistor is known, Ohm's law can again be used to find the current in the second resistor. The total power consumed by the parallel combination can be found calculating the power consumed by each resistor from Equation.6b: P = I R. Then, the total power consumed is the sum of the power consumed by each resistor. SOLUTION Using data for the second resistor, the voltage across the resistors is equal to V = IR = (3. A)(64. Ω)= 19V a. The current through the 4.-Ω resistor is I = V R = 19V 4. Ω = 4.57 A b. The power consumed by the 4.-Ω resistor is P = I R = (4.57 A) (4. Ω) = 877 W while the power consumed by the 64.-Ω resistor is P = I R = (3. A) (64. Ω ) = 576 W Therefore the total power consumed by the two resistors is 877 W W = 145 W.

7 55. SSM REASONING The equivalent resistance of the three devices in parallel is R p,and we can find the value of R p by using our knowledge of the total power consumption of the circuit; the value of R p can be found from Equation.6c, P = V / R p. Ohm's law (Equation., V = IR) can then be used to find the current through the circuit. SOLUTION a. The total power used by the circuit is P = 165 W + 19 W +15 W = 399 W. The equivalent resistance of the circuit is R p = V P = (1 V) 399 W = 3.6 Ω b. The total current through the circuit is I = V R p = 1 V 3.6 Ω = 33 A This current is larger than the rating of the circuit breaker; therefore, the breaker will open.

8 58. REASONING We will approach this problem in parts. The resistors that are in series will be combined according to Equation.16, and the resistors that are in parallel will be combined according to Equation.17. SOLUTION The 1. Ω,. Ω and 3. Ω resistors are in series with an equivalent resistance of R s =6.Ω.. Ω 6. Ω 4. Ω 3. Ω 6. Ω This equivalent resistor of 6. is in parallel with the 3.-Ω resistor, so = + R 6. Ω 3. Ω P. Ω 6. Ω 4. Ω. Ω R P =. Ω This new equivalent resistor of. is in series with the 6.- Ω resistor, so R s '=8.Ω.. Ω 4. Ω 8. Ω R s ' is in parallel with the 4.-Ω resistor, so = + R 8. Ω 4. Ω P. Ω.67 Ω R =.67 Ω P Finally, R p ' is in series with the.-ω, so the total equivalent resistance is 4.67.

9 6. REASONING The circuit diagram is shown at the 6. Ω right. We can find the current in the 1.-Ω resistor by using Ohm s law, provided that we can obtain a value for V AB, the voltage between points A and B in the. Ω A B diagram. To find V AB, we will also apply Ohm s law, 1. Ω this time by multiplying the current from the battery times R AB, the equivalent parallel resistance between A 15. V and B. The current from the battery can be obtained by applying Ohm s law once again, now to the entire circuit and using the total equivalent resistance of the series combination of the.-ω resistor and R AB. Once the current in the 1.-Ω resistor is found, the power dissipated in it can be obtained from Equation.6b, P = I R. SOLUTION a. According to Ohm s law, the current in the 1.-Ω resistor is I 1 = V AB /(1. Ω). To find V AB, we note that the equivalent parallel resistance between points A and B can be obtained from Equation.17 as follows: or R 4. AB R = 6. Ω + 1. Ω = Ω AB This resistance of 4. Ω is in series with the.-ω resistance, so that, according to Equation.16, the total equivalent resistance connected across the battery is 4. Ω +. Ω = 6. Ω. Applying Ohm s law to the entire circuit, we can see that the current from the battery is I = 15. V 6. Ω =.5 A Again applying Ohm s law, this time to the resistance R AB,wefindthat V AB ( ) R ( )( ) =.5 A =.5 A 4. Ω = 1. V Finally, we can see that the current in the 1.-Ω resistor is AB V 1. V 1 Ω 1 Ω AB I = = = A b. The power dissipated in the 1.-Ω resistor is given by Equation.6b as

10 ( ) ( ) P= I R= A 1. Ω =.833 W 1

11 64. REASONING The power P dissipated in each resistance R is given by Equation.6b as P= I R,whereI is the current. This means that we need to determine the current in each resistor in order to calculate the power. The current in R 1 isthesameasthecurrentinthe equivalent resistance for the circuit. Since R and R 3 are in parallel and equal, the current in R 1 splits into two equal parts at the junction A in the circuit. SOLUTION To determine the equivalent resistance of the circuit, we note that the parallel combination of R and R 3 is in series with R 1. The equivalent resistance of the parallel combination can be obtained from Equation.17 as follows: or RP 88 R = P 576 Ω Ω = Ω This 88- resistance is in series with R 1, so that the equivalent resistance of the circuit is given by Equation.16 as R eq = 576 Ω+ 88 Ω= 864 Ω To find the current from the battery we use Ohm s law: I V 1. V = = =.139 A R 864 Ω eq Since this is the current in R 1, Equation.6b gives the power dissipated in R 1 as ( ) ( ) P1 = I1R1 =.139 A 576 Ω = 11.1 W R and R 3 are in parallel and equal, so that the current in R 1 splits into two equal parts at the junction A. As a result, there is a current of 1 (.139 A ) in R and in R 3. Again using Equation.6b, we find that the power dissipated in each of these two resistors is 1 IR 1 3 I3R 3 ( ) ( ) P = =.139 A 576 Ω =.78 W ( ) ( ) P = =.139 A 576 Ω =.78 W