A commercial resistor can safely dissipate power only up to a certain rated value. Beyond this value, the resistor becomes excessively hot and often cracks apart. The resistor used in this experiment was rated by the manufacturer to be a 2 watt resistor. What is the largest voltage that can safely be applied across the resistor? Hint: The power P dissipated by a resistor R is given by P = I2R = V2R = IV where I is the current and V is the voltage.

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A 120.0 ohm resistor, a 60.0 ohm resistor, and a 40.0ohm resistor are connected in parallel and placed across a 12.0-V battery. What is the current of the entire circuit?

a.
0.800 A
c.
0.600 A
b.
1.26 A
d.
0.062 A

To find the largest voltage that can be safely applied across the resistor, we need to use the power rating and the given formula:

P = V^2 / R

where P is the power dissipated by the resistor, V is the voltage, and R is the resistance.

Since we are given that the resistor is rated at 2 watts, we can substitute this value into the equation:

2 = V^2 / R

To find the largest voltage, we need to determine the highest possible value of V that satisfies this equation.

Since we know the power rating and the resistance is not given, we can use the formula:

P = I^2 * R

We can substitute this into the power equation:

2 = (V^2 / R) * R

2 = V^2

Taking the square root of both sides:

√2 = V

Therefore, the largest voltage that can be safely applied across the resistor is approximately 1.41 volts.