Electric Circuits - Sample Problems with Solutions | PHYS 122, Study notes of Physics

Download Electric Circuits - Sample Problems with Solutions | PHYS 122 and more Study notes Physics in PDF only on Docsity! CHAPTER 20|ELECTRIC CIRCUITS PROBLEMS saggy sea ie RST SERRE Sae (SsM] REASONING ‘Since current is defined as charge per unit time, the current used by the portable compact disc player is equal to the charge provided by the battery pack (180 C) divided by the time in which the charge is delivered -(2.0 h). SOLUTION The amount of current that the player uses in operation is determined from Equation 20.1: _ Ag 180C (10h “at 20h 3600 oe Converts hours to seconds =| 0.025 A i REASONING The current J is defined in Equation 20.1 as the amount of charge Ag per unit of time Af that flows in a wire. Therefore, the amount of charge is the product of the current and the time interval. The number of electrons is equal to the charge that flows divided by the magnitude of the charge on an electron. SOLUTION a. The amount of charge that flows is Aq =1At =(18 A)(2.0x 10°s)= 3:6x107 C b. The number of electrons N is equal to the amount of charge divided by e, the magnitude of the charge on an electron. 2 yw Bd 2 3:8%107 © _ 13a" e 1,60x10"C REASONING AND SOLUTION We know that V = IR. Therefore, 4. REASONING AND SOLUTION Ohm’s law gives _¥_ ov _ “R 149 — 5. REASONING AND SOLUTION Ohm's law (Equation 20.2, V = IR) gives the result directly J=—=—— =[22A 6. REASONING AND SOLUTION First determine the total charge delivered to the battery using Equation 20.1: Aq =IAt = (6.0 A)(5.0 h)[(600 s)/(1 h)] = 1.1 x 10°C To find the energy delivered to the battery, multiply this charge by the energy per unit charge (i.e., the voltage) to get Energy = (Aq)V = (1.1 x 10° C)(12 V) = [1.3 x108J ee 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: 1kw 1000 W Energy used per set = Pt = (75 w){ Joo h) (6.10b) The total cost of operating the TV sets is 1000 W 1kW-h Total cost = (110 million se) |( w)( 1kW Jeo | $0.10 } =|$5.0x 10° 25. REASONING AND SOLUTION According to Equation 20.6a, we know P = IV, so that I= P/V = (140 W)(120 V) = 26. REASONING AND SOLUTION The power delivered is P = VI so a. Pyg = Vig = (120 V)(11 A) = [1300 W) b. Pye = Wij, = (120 V)(4.0 A) = [480 W, c. The energy is E = Pt so, Epd Eve = Pratya/(P ycty,) = (1300 W)(15 min)/[(480 W)30.0 min)] = 39. {ssm] REASONING The equivalent series resistance R, is the sum of the resistances of the three resistors. The potential difference V can be determined from Ohm's law as V = IR,. SOLUTION a. The equivalent resistance is R, =25 2445 2475 O=[745 8] b. The potential difference across the three resistors is V= IR, =(0.51 A145 Q)=[74V 40. 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 = IR, =1(R, +R) = (0.12 A)(472+280)=[9.0V 41. REASONING AND SOLUTION The equivalent resistance of the circuit is R,=R, +R, =36.02+ 1800=5400 Ohm's law for the circuit gives I= V/R, = (15.0 VV(54.0 Q) = 0.278 A a. Ohm's law for R, gives V, = (0.278 A)(36.0 Q)= b. Ohm's law for R, gives V, = (0.278 A)(18.0 Q) = 42. REASONING - According to Equation 20.2, the resistance R of the resistor is equal to the voltage Vp across it divided by the current J, or R = Vg Al. Since the resistor, the lamp, and the voltage source are in series, the voltage across the resistor is Va = 120.0 V - V,, where V,, is the voltage across the lamp. Thus, the resistance is _120.0V-Y, - T R Since V, is known, we need only determine the current in the circuit. Since we know the voltage Vj, across the lamp and the power P dissipated by it, we can use Equation 20.6a to find the current: J = P/V, . The resistance can be written as . 53. The resistance Rj 9 of the 100.0-W filament is 2 2 2 0200" ras R = 100" P 100.0 W . [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 20.2: V = JR). 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 20.6b: P= PR. 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=(3.00 A)(64.0 2)= 1920 a. The current through the 42.0-2 resistor is Ve 192V [=a =—— =| 457A R 42.09 b. The power consumed by the 42.0-Q resistor is P=I? R=(4.57 A)? (42.0 Q)= 877W while the power consumed by the 64.0-Q resistor is P=I? R=(3.00 A)?(64.0 Q)= 576 W Therefore the total power consumed by the two resistors is 877 W +576 W =| 1450 W| . . REASONING AND SOLUTION Each piece has a resistance of 1/3 R. Then UR, = VB R)+ 1/3 R)+ 14/3 R)=9R oor R= REASONING The total power is given by Equation 20.15¢ as P = ve. / R,, where R, is the equivalent parallel resistance of the heater and the lamp. Since the total power and the rms voltage are known, we can use this expression to obtain the equivalent parallel resistance. This equivalent resistance is related to the individual resistances of the heater and the lamp via Equation 20.17, which is R;' Ramp can be found once R, is known, 4 -1 : boos = Ricater + Riamp - Since Rreater is given, SOLUTION According to Equation 20, 15c, the equivalent parallel resistance is R= vulg% Using this result in Equation 20.17 gives =42x103 g-1 owt 1 Ry Vins IP Rrcatet Riamp Rearranging this expression shows that 1 PF 1 ww Ramp v2. * Roester (120 vy? 6.0107 9 Therefore, . 1 Ramp =F =| 240 vam "4.2.x 102 Qt 54. REASONING The electric heater and the toaster are connected in parallel with the voltage source, so each receives the same voltage as the source. The is given by Equation 20,14, Tans = Vimig/Rs where Vins tms-current through the toaster is the rms-voltage across the toaster and R is its resistance. The total power ‘supplied to the heater and toaster is given by Equation 20.15¢ as P= ve /R,, where Rp-is the equivalent resistance of the parallel circuit. SOLUTION a, The voltage across the heater is the same as that of the generator, V1. = b. The current through the toaster is Vow _ 120.0V 1 =—Bi = =[7.06 A) m Reaser 17.02 120.0 Vj. (20.14) c. The average power supplied to the héater and toaster is P= ve 7 Ry. The: equivalent resistance can be obtained from Equation 20.17: 1 1 + 1 Ry Reeser Reser The average power becomes ~ k 2 1 1 2f 1 1 P=M=Ve, + = (120.0 V) | ——— +> = |= [2350 W Rp Reeser Roaster 9602 17.02 - [SSM REASONING The equivalent resistance of the three devices in parallel is Rp, and we can find the value of R, by using our knowledge of the total power consumption of the circuit; the value of Ry can be found from Equation 20.6c, P= vr I Rp: Ohm's law (Equation 20.2, V = IR) can then be used to find the current through the circuit. SOLUTION a, The total power used by the circuit is P = 1650 W + 1090 W +1250 W= 3990 W. The equivalent resistance of the circuit is 2 2 2 OW _Gea P“"p ~ 3990 W b. The total current through the circuit is p - 20Y Ga R, 3.62 This current is larger than the rating of the circuit breaker; therefore, the breaker will open |. Veapesr ay rivurguiD 707 40 Roa4s = 2R/5, The resistor R, is in series with R345, and the equivalent resistance of this combination is the equivalent resistance of the circuit. Thus, we have 2R _7R Rosas = B+ Rings = Ra power delivered to the circuit is v? v? mG jolving for the resistance R, we find that P= 2 2 p= UO) ose 7P 7(58 W)  "8. REASONING AND SOLUTION. Let the current through the 20.0 Q be I, and flow to the right. _ Let the current through the 10.0 Q’be I, and flow up. © | Let the current through the 5.0 Q be I, and flow to the tight. Applying the loop rule to the left loop gives 20.0.1, - 10,01, =0 and to the right loop 10.0 1, + 5.01, = 30.0 The junction rule applied to the upper junction gives 1, +1,-1,=0 A simultaneous solution of the above gives 1, = [1,71 Aj. pei Since the answer is positive, the current flows in the assumed direction. ‘That is, from the bottom to the top] .