A car has a 12 volt battery. The engine has a resistance of 0.22 ohms. How many amps will be drawn from the battery when the key is turned?
V=IR
V= 12v
R= 0.22 ohms
I = V/R
= 12/0.22
=54.5 amps
E= I x R is the formula
the answer is .5454
I=amperes
R=ohms
E=volts
12volts/22
actually David is right. I forgot the decimal place from the 22 so it is 54.5 amps
thank you
Three cells are connected. The external resistance of each cell is 2 ohms. The voltage of each cell is 1.5 volts. The resistance of the external circuit is 2.3 ohms. What is the circuit current to the nearest tenth ampere?
To calculate the amperage drawn from the battery, we can use Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R).
The given voltage is 12 volts, and the resistance of the engine is 0.22 ohms. Therefore, we can use the formula:
I = V / R
Substituting the values:
I = 12 volts / 0.22 ohms
Now let's calculate the amperage drawn from the battery.
I ≈ 54.55 amps
So, when the key is turned, approximately 54.55 amps will be drawn from the battery. Keep in mind that this calculation assumes ideal conditions and does not take into account other factors such as inefficiencies or other components in the circuit.