Prettyluhjasminidn
Answered

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Use the equation to answer the question.

[tex]\[ I = \frac{E}{R} \][/tex]

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?

[tex]\[ I \text{ (to the nearest hundredth) } = \boxed{\ \ \ } \text{ amperes.} \][/tex]

Sagot :

GinnyAnsweridn
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]
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