Question
According to Ohm’s Law, voltage is directly proportional to the strength of current multiplied by resistance within a circuit (V = I x R). Given this, if a resistor in a circuit has a value of 550 Ω (Ohms) and is powered by a 9 V (volt) battery, what is the current through the circuit?
Group of answer choices
4,950 A (amps)
0.0164 A (amps), or 16.4 mA (milliamps)*
61.1 A (amps)
Group of answer choices
4,950 A (amps)
0.0164 A (amps), or 16.4 mA (milliamps)*
61.1 A (amps)
Answers
GPT-4o mini
To find the current (I) through the circuit using Ohm's Law (V = I x R), you can rearrange the formula to solve for current:
\[ I = \frac{V}{R} \]
Given:
- Voltage (V) = 9 volts
- Resistance (R) = 550 ohms
Now, substitute the known values into the formula:
\[ I = \frac{9 \, \text{V}}{550 \, \Omega} \]
Calculating this gives:
\[ I = \frac{9}{550} \approx 0.01636 \, \text{A} \]
This is approximately 0.0164 A, which is equivalent to 16.4 mA (milliamps).
Therefore, the correct answer is:
**0.0164 A (amps), or 16.4 mA (milliamps)**.
\[ I = \frac{V}{R} \]
Given:
- Voltage (V) = 9 volts
- Resistance (R) = 550 ohms
Now, substitute the known values into the formula:
\[ I = \frac{9 \, \text{V}}{550 \, \Omega} \]
Calculating this gives:
\[ I = \frac{9}{550} \approx 0.01636 \, \text{A} \]
This is approximately 0.0164 A, which is equivalent to 16.4 mA (milliamps).
Therefore, the correct answer is:
**0.0164 A (amps), or 16.4 mA (milliamps)**.