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.

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 =  6Ω  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