In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law the current t

Question

In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law the current through the circuit is _____________ Amps. If a second identical lamp is connected in series, the 6-volt battery must push a charge through a total resistance of ________ Ω. The current in the circuit is then __________ Amps. If a third identical lamp is connected in series, the total resistance is now _________Ω. The current through all three lamps in series is now _________ Amps. The current through each individual lamp is __________ Amps. What is the power when a voltage of 120 volts drives a current of 3 amps through a device? What is the current when a 90-W light bulb is connected to 120 V? How much current does a 75-W light bulb draw when connected to 120 V? If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it? If a 60 W light bulb at 120 V is left on in your house to prevent burglary, and the power company charges 10 cents per kilowatt-hour, how much will it cost to leave the bulb on for 30 days? Show your work.

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Nho 6 months 2021-08-05T17:50:08+00:00 1 Answers 18 views 0

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    2021-08-05T17:51:25+00:00

    Answer:

    The answers and workings is in the Explanation section

    Explanation:

    In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law the current through the circuit is _____________ Amps.

    According to Ohm’s law V =I*R  

    Where V = Voltage, I = Current and R = Resistance

    I = V/R =6/3 =2 Amps of current

    Answer = 2 Amps

    If a second identical lamp is connected in series, the 6-volt battery must push a charge through a total resistance of ________ Ω.  The current in the circuit is then __________ Amps.

    Since the second lamp is connected in series with the first, the total resistance will R₂ = 3 + 3 = 6Ω

    2 resistance in series  R₂ =  6Ω

    The current in the circuit with the two lamps connected in series is I₂ =V/R₂ =6/6 = 1 Amps

    The current is 1 Amps

    Answer =    and 1 Amps

    If a third identical lamp is connected in series, the total resistance is now _________Ω.  The current through all three lamps in series is now _________ Amps.  

    Since the third lamp is connected in series with the first and second, the total resistance will R₃ = 3 + 3 + 3 = 9Ω

    total resistance of the 3 lamps R₃ =  9Ω

    The current in the circuit with the three lamps connected in series is

    I =V/R₃ =6/9 = 0.67 Amps

    The current through the 3 lamps I₃ = 0.67 Amps

    Answer =  9Ω  and 0.67 Amps

    The current through each individual lamp is __________ Amps.  

    Since all 3 lamps are connected in series, the same current will flow through each of the  3 lamps, and that current is I₃  

    The current through each individual lamp is 0.67 Amp

    Answer = 0.67 Amp

    What is the power when a voltage of 120 volts drives a current of 3 amps through a device?  

    The formula for power P = I*V =120*3 = 360 Watts

    power P  = 360 Watts

    Answer = 360 Watts

    What is the current when a 90-W light bulb is connected to 120 V?

    From P =I*V, make I the subject of the formula, I = P/V =90/120 = 0.75


    Current= 0.75 Amps

    Answer =  0.75 Amps

    How much current does a 75-W light bulb draw when connected to 120 V?  

    Current I =P/V = 75/120 = 0.625 Amps.

    Answer = 0.625 Amps

    If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it?

    Voltage V =P/I =6/3 =2 Volts

    Answer = 2 Volts

    If a 60 W light bulb at 120 V is left on in your house to prevent burglary, and the power company charges 10 cents per kilowatt-hour, how much will it cost to leave the bulb on for 30 days? Show your work.

    24 hours make 1 day, so the number of hours the bulb was left on = 24 *30 = 720 hours

    The power rating of the bulb is 60W = 60/1000 = 0.06 KiloWatt

    Total power consumed in kilowatt-hour = 0.06 * 720 = 43.2 kilowatt-hour

    Cost for 30 days = 0.1*43.2 = $4.32 ( note that 10 Cents = $0.1)

    Answer =  $4.32

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