Question

1.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 2 Amps.2.If a second identical lamp is connected in series, the 6-volt battery must push a charge through a total resistance of 6Ω. The current in the circuit is then 1 Amps.3.If a third identical lamp is connected in series, the total resistance is now 9Ω.4.The current through all three lamps in series is now _________ Amps. The current through each individual lamp is __________ Amps.

Answer 1

The current through all three lamps in series is now 0.67 Amps. The current through each individual lamp is 0.67 Amps.

Step-by-step explanation

There are three lamps connected in series, each with a resistance of 3 Ω, resulting in a total resistance of 9 Ω.

By Ohm's law, if the voltage from the battery is 6 V, then the current through all three lamps - i.e. the total current in the circuit is,

`I=(V)/(R_(eq))=(6V)/(9\Omega)=(2)/(3)Amps\approx0.67Amps`

And, since the three lamps are connected in series - which means there are no dividing paths, the current through each individual lamp is the same as the total current of the circuit, 0.67 Amps.

Hence, the current through all three lamps and through each individual lamp is 0.67 Amps, rounded to the nearest hundredth.