Certainly! Let's solve this step-by-step using the given equation [tex]\( I = \frac{E}{R} \)[/tex].
We are given:
- [tex]\( E \)[/tex] (Voltage) = 12 volts
- [tex]\( R \)[/tex] (Resistance) = 0.22 ohms
We need to find the current [tex]\( I \)[/tex] in amperes, which can be calculated using the formula:
[tex]\[ I = \frac{E}{R} \][/tex]
Now, let's substitute the given values into the equation:
[tex]\[ I = \frac{12 \text{ volts}}{0.22 \text{ ohms}} \][/tex]
This division gives us:
[tex]\[ I \approx 54.54545454545455 \text{ amperes} \][/tex]
To express this current to the nearest hundredth, we round 54.54545454545455 to:
[tex]\[ I \approx 54.55 \text{ amperes} \][/tex]
Therefore, the current drawn from the battery when the key is turned is:
[tex]\[ I = 54.55 \text{ amperes} \][/tex]
