If a current of 2 amperes flows through a 50-ohm resistor, what is the voltage across the resistor?

If a current of 2 amperes flows through a 50-ohm resistor, what is the voltage across the resistor?

• A. 50 volts
• B. 100 volts
• C. 150 volts
• D. 200 volts

Answer: (B)

To determine the voltage across a resistor when a current flows through it, you can apply Ohm's Law, which is stated as: \[ V = I \times R \] where \( V \) is the voltage across the resistor, \( I \) is the current flowing through it, and \( R \) is the resistance in ohms. In this scenario, the current \( I \) is 2 amperes and the resistance \( R \) is 50 ohms. Plugging these values into the formula gives: \[ V = 2 \, \text{A} \times 50 \, \Omega = 100 \, \text{V} \] This means the voltage across the resistor is indeed 100 volts. Therefore, the correct answer reflects the application of Ohm's Law correctly. Regarding other choices, it's important to note how they compare when applying the same formula with different assumptions about current or resistance. The incorrect values arise from either miscalculating the multiplication or misunderstanding the relationship defined by Ohm’s Law. In practical applications, recognizing how to use the formula accurately is critical for circuit analysis and troubleshooting.

To determine the voltage across a resistor when a current flows through it, you can apply Ohm's Law, which is stated as:

[ V = I \times R ]

where ( V ) is the voltage across the resistor, ( I ) is the current flowing through it, and ( R ) is the resistance in ohms.

In this scenario, the current ( I ) is 2 amperes and the resistance ( R ) is 50 ohms. Plugging these values into the formula gives:

[ V = 2 , \text{A} \times 50 , \Omega = 100 , \text{V} ]

This means the voltage across the resistor is indeed 100 volts. Therefore, the correct answer reflects the application of Ohm's Law correctly.

Regarding other choices, it's important to note how they compare when applying the same formula with different assumptions about current or resistance. The incorrect values arise from either miscalculating the multiplication or misunderstanding the relationship defined by Ohm’s Law. In practical applications, recognizing how to use the formula accurately is critical for circuit analysis and troubleshooting.