what is the power dissipated by a 10 ohm resistor through which a 2 ampere current is flowing?
V = I x R
P = V x I
P = R x I^2
P = 10 x 2^2 = 40 watts = 136.5 btu/hr
Well, if we do the math, the power dissipated by the resistor can be calculated using the formula P = I^2 * R, where P is the power, I is the current, and R is the resistance. So, in this case, we have 2^2 * 10, which equals 40 watts. That's enough power to make a resistor blush!
To calculate the power dissipated by a resistor, we can use the formula:
Power (P) = Current (I) x Current (I) x Resistance (R)
In this case, the current flowing through the resistor is given as 2 amperes (I), and the resistance of the resistor is given as 10 ohms (R).
Therefore, we can substitute these values into the formula to find the power dissipated:
P = 2 A x 2 A x 10 Ω
Simplifying the equation:
P = 4 A² x 10 Ω
P = 40 A²Ω
The unit of power is watts (W). Therefore, the power dissipated by the 10 ohm resistor with a 2 ampere current flowing through it is 40 watts (W).
To calculate the power dissipated by a resistor, you can use the formula:
Power (P) = Current (I) * Current (I) * Resistance (R)
Given:
Current (I) = 2 amperes
Resistance (R) = 10 ohms
Substituting these values into the formula, we get:
P = 2 A * 2 A * 10 Ω
Simplifying the equation:
P = 4 A^2 * 10 Ω
P = 40 A^2 Ω
Therefore, the power dissipated by the 10-ohm resistor with a 2-ampere current flowing through it is 40 watts.