Current Electricity

626 18) A 150 * branch in a circuit must be reduced to 93 * A resistor will be added to this branch of the circuit to make this change. What value of resistance should be used and how must the resistor be connected?

(150*93)/(150-93)= 244.7 * Parallel

600 15) Draw a circuit using a battery, a lamp, a potentiometer to adjust the lamp's brightness, and an on-off switch?

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600 16) Repeat (Draw a circuit using a battery, a lamp, a potentiometer to adjust the lamp's brightness, and an on-off switch?) adding ammeter and voltmeter across the lamp

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605 31) Rework the last problem by assuming that the battery requires the application of 14 V when it is recharging. (An automotive battery can deliver 55 A at 12 V for 1 h and requires 1.3 times as much energy for recharge due to its less than perfect efficiency. How long will it take to charge the battery using a current of 7.5 A? Assume that the charging voltage is the same as the discharging voltage.)

1.3*i*v*t 1.3*55*12*1=858 T=E/(iV) or 858/(7.5*14) or 8.2 h

619 2) A 10 *, 15 * and 5 * resistors are connected in a series circuit with a 90 V battery. What is the equivalent resistance of the circuit? What is the current in the circuit?

10+15+5=30; 90/30=3 A

622 10) A series circuit is made up of a 12 V battery and three resistors. The voltage across one resistor is 1.21 V, and the voltage across another resistor is 3.33 V. What is the voltage across the third resistor?

12-1.21-3.33= 7.46V

630 25) A series parallel circuit has three resistors; one dissipates 2 W, the second 3 W and the third 1.5 W. How much current does the circuit require from a 12 V battery?

2+1.5+3=6.5 P=IV 6.5=I(12) I=0.54

619 1) Three 20 * resistors are connected in series across a 120 V generator. What is the equivalent resistance of the circuit? What is the current in the circuit?

20+20+20=60 A V=IR 120=I(60) I=2 A

623 13) A student makes a voltage divider from a 45 V battery, a 475 k* resistor and a 235 k* resistor. The output is measured across the smaller resistor. What is the voltage?

235+475= 710 45=710I I=6.3*10^-5 A V=IR (6.3*10^-5)*(235)= 14.9V

619 5) Calculate the voltage drops across the three resistors in problem 2 (3A and 90 V; 10, 15, 5 R), and verify that their sum equals the voltage of the battery.

3(10)+3(15)+3(5)=90 V

636 53) If you have a 6 V battery and many 1.5 V bulbs how could you connect them so that they light but do don't have more than 1.5 V across each bulb?

6/1.5=4 bulbs

636 51) What happens to the current in the other tow lamps if one lamp in a three parallel circuit burns out?

Current and brightness will drop to 0

603 27) A 120 V water heater takes 2.2 h to heat a given volume of water to a certain temp. How long would a 240 V unit operating with the same current take to accomplish the same task?

E=IVt I(2V)(t/2) Doubling voltage will divide time by 2 2.2/2= 1.1 h

605 30) An automotive battery can deliver 55 A at 12 V for 1 h and requires 1.3 times as much energy for recharge due to its less than perfect efficiency. How long will it take to charge the battery using a current of 7.5 A? Assume that the charging voltage is the same as the discharging voltage.

Echarge=1.3*I*v*t 1.3*55*12*1 =858 T=E/(IV) or 858/(7.5*12) or 9.5 h

610 54) Describe two ways to increase the current in a circuit?

Increase the voltage while holding resistance constant Decreasing the resistance while holding voltage constant

636 52) An engineer needs a 10 * resistor and a 15 * resistor but there are only 30 * resistors in stock. Must new resistors be purchased?

No

594 5) A flashlight bulb is rated 0.90 W. If the light bulb drops 3 V, how much current goes through it?

P=IV or 0.9=I(3) or 0.9/3 =0.3 A

594 2) A car batter causes a current of 2 A through a lamp and produces 12 V across it. What is the power used by the lamp?

P=IV or 2(12) =24 W

630 26) There are 11 lights in series, and they are in series with two lights in parallel. If the 13 lights are identical, which will burn the brightest?

The 11 lights will burn brighter. The parallel will conduct half of the current of the series lights and they will burn at one fourth the intensity of the series lights since P=I^2R

623 14) Select a resistor to be used as part of a voltage divider along with a 1.2k* resistor. The drop across the 1.2 k* resistor is to be 2.2 V when the supply is 12 V.

V=IR 12=1200I; I=.01 A V=(12v-2.2v); V=9.8v V=IR 9.8=.01R R=980

598 8) A sensor uses 2.0*10^-4 A of current when it is operated by a 3 V battery. What is the resistance of the sensor circuit?

V=IR 3=(2*10^-4)(R) or 3/(2*10^-4) =1.5*10^4

600 13) Draw a series-circuit diagram showing a 4.5 V battery, a resistor, and an ammeter that reads 85 mA. Determine the resistance and label the resistor. Choose a direction for the conventional current and indicate the positive terminal of the battery.

V=IR 4.5=(.085)(R) or 4.5/.085 53 Resistance

598 6) An automobile panel lamp with a resistance of 33 is placed across a 12 V battery. What is the current through the circuit?

V=IR or 12=I(33) or 12/33 =0.36 A

598 7) A motor with an operating resistance of 32 is connected to a voltage source. The current in the circuit is 3.8 A. What is the voltage of the source?

V=IR or 32(3.8) =1.2*10^2

600 12) Draw a circuit diagram to include a 60 V battery, an ammeter, and a resistance of 12.5 in series. Indicate the ammeter reading and the direction of the current?

V=IR or 60=I(12.5) I=60/12.5 or 4.8 A

610 52) Explain why a cow experiences a mild shock when it touches an electric fence?

When the cow comes into contact with the fence, it becomes an electric ground which sends an electric current through the cow, the pain is experienced through the current that flows through the cow

598 11) A resistor is added to the lamp in the previous problem to reduce the current to half of its original value. (original I=0.6 and original R=2.1*10^2; V=125) a) What is the potential difference across the lamp? b) How much resistance was added to the circuit? c) How much power is now dissipated in the lamp?

a) 0.3/2= 0.6 V=IR or V=(0.3)(2.1*10^2) =6.3*10^1 V b) V=IR; V=125 125=0.3R or 125/0.3 or 4.2*10^2 (4.2*10^2)-(2.1*10^2) =2.1*10^2 c) P=IV (0.3)(6.3*10^1) =19 W

626 16) A 120 * resistor, a 60 resistor *, and a 40 resistor* are connected in parallel and placed across a 12 V battery. a) what is the equivalent resistance of the parallel circuit? b) what is the current through the entire circuit? c) what is the current through each branch of the circuit?

a) 1/120+1/60+1/40= 20 * b) V=IR 12=20I; I=.6 c) 12/120=.1 ; 12/60=.2 ; 12/40=.3

626 17) Suppose that one of the 15 * resistors in problem 15 is replaced by a 10 * resistor. (120, 60, 40 resistor, 12 V) a) does the equivalent resistance change? If so how? b) Does the amount of current through the entire circuit change? If so, in what way? c) does the amount of current through the other 15 * resistors change? If so, how?

a) 1/15+1/15+1/10 R=4.28, other was 5 * b) V=IR 30=4.28I; I=7 other was 6A c) No the additional current only goes through the 10 *

603 25) A 100 W lightbulb is 22 percent efficient. This means that 22 percent of the electric energy is converted to light energy a) how many joules does the lightbulb convert into light each minute it is in operation b) how many joules of thermal energy does the lightbulb produce each minute

a) 100(60) =6000 J b) 6000*.22 =1320 J into heat

619 4) A string of holiday lights has ten bulbs with equal resistance connected in series. When the string of lights is connected to a 120 V outlet, the current through the bulbs is 0.06 A. a. what is the equivalent resistance of the circuit? b. what is the resistance of each bulb?

a) 120/0.06 =2000 b) 2000/10= 200

626 15) Three 15 * are connected in parallel and placed across a 30 V battery. a) what is the equivalent resistance of the parallel circuit b) what is the current through the entire circuit? c) what is the current through each branch of the circuit?

a) 15/3= 5 b) V=IR; 30=I(5); I=6 c) 15, and are 3 gets 1/3 or 6/3 or 2A

623 11) A 22 * resistor and a 33 * resistors are connected in series and placed across a 120 V potential difference. a) what is the equivalent resistance of the circuit? b) what is the current in the circuit? c) what is the voltage drop across each resistor? d) Find the total voltage drop across the three resistors?

a) 22+33= 55 b)120/5.5= 2.2 A c) 48 V, 72 V **** d) 1.20*10^2

598 10) A 75 W lamp is connected to 125 V. a) what is the current through the lamp? b) what is the resistance of the lamp?

a) P=IV 75=I(125) or 75/125 =0.6 b) V=IR 125=0.6(R) or 125/0.6 =2.1*10^2

603 24) A 36 * resistor is connected across a 45 V battery a) what is the current in the circuit b) how much energy is used by the resistor in 5 min?

a) V=IR 45=I(36) or 45/36 =1.2 A b) E=(v^2/r)*t (45V)^2/39)*(5*60) 1.6*10^4 J

622 7) Suppose the circuit shown in Example Problem 1 has these values: Ra=255 *, Rb=292 *, and Va=17.0 V. No other information is available. a) what is the current in the circuit? b) what is the battery voltage? c) what are the total power dissipation and the individual power dissipations? d) does the sum of the individual power dissipations in the circuit equal the total power dissipation in the circuit? Explain?

a) V=IR 17=I(255) =.066 A b) V=IR .066*547= 36.5 c) (.066)*17=2.43 (.066)*(35-17)=1.13 (.066)*35=1.23 =about 2.43 d) 2.43=2.43+1.13+1.23

605 29) A digital clock has a resistance of 12,000 * and is plugged into a 115 V outlet. a) how much current does it draw? b) how much power does it use? c) If the owner of the clock pays 0.12 per kWh, how much does it cost to operate the clock for 30 days?

a) V=IR or 115=I(12000) or 115/12000 or I= 9.6*10^-3 A b) P=IV or (9.6*10^-3)*115 or 1.1 W c) (1.1*10^-3 kWh)*0.12*(30*24) or 0.10

619 3) A 9V battery is in a cicuit with three resistors connected in a series. a. If the resistance of one the resistors increases, how will the equivalent resistance change? b. what will happen to the current? c. Will there be any change in the battery voltage?

a) increases b) If it increases then current decreases c) No, it stays the same

623 12) Three of 3.3 k*, 4.7 k*, and 3.9 k* are connected in series across a 12 V battery. a) what is the equivalent resistance? b) what is the current through the resistor? c) what is the voltage drop across each resistor? d) find the total voltage drop across the three resistors?

a) k=1000* so that 3300+4700+3900= 11900 b) V=IR 12/11900= .001A c) (.001)(3300)=3.3V (.001)(4700)=4.7V (.001)(3900)= 3.9V d) 3.3+4.7+3.9= 12 V

636 54) Two lamps have different resistances, one larger than the other. a) if the lamps are connected in parallel, which is brighter (dissipates more power)? b) when the lamps are connected in series, which lamp is brighter?

a) less resistant one in parallel will be brighter b) series higher resistant one will be brighter