Using Ohm's law, which formula represents power dissipated by a resistor?

Using Ohm's law, which formula represents power dissipated by a resistor?

• A. P = IV
• B. P = I²R
• C. P = V²/R
• D. All of the above

Answer: (D)

The power dissipated by a resistor can be represented using multiple equations derived from Ohm's law, which states that Voltage (V) = Current (I) x Resistance (R). Each of the options provided reflects a correct relationship involving power (P), current (I), voltage (V), and resistance (R). One fundamental formula for power is P = IV, indicating that the electrical power is the product of current flowing through a circuit and the voltage across the resistor. Additionally, by substituting Ohm's law into this equation, you can derive other expressions for power. From P = IV, if you substitute for voltage (V = IR), the equation becomes P = I(IR), which simplifies to P = I²R. This shows how current squared times resistance also gives power. Similarly, you can manipulate the original formula P = IV to derive another relationship. If you substitute for current (I = V/R), the equation becomes P = (V/R)V, which simplifies to P = V²/R. This indicates that voltage squared divided by resistance provides another expression for power. Thus, all three formulas are valid expressions for calculating power in a resistor, reinforcing the idea that multiple relationships exist to define how power, voltage, current

The power dissipated by a resistor can be represented using multiple equations derived from Ohm's law, which states that Voltage (V) = Current (I) x Resistance (R). Each of the options provided reflects a correct relationship involving power (P), current (I), voltage (V), and resistance (R).

One fundamental formula for power is P = IV, indicating that the electrical power is the product of current flowing through a circuit and the voltage across the resistor.

Additionally, by substituting Ohm's law into this equation, you can derive other expressions for power. From P = IV, if you substitute for voltage (V = IR), the equation becomes P = I(IR), which simplifies to P = I²R. This shows how current squared times resistance also gives power.

Similarly, you can manipulate the original formula P = IV to derive another relationship. If you substitute for current (I = V/R), the equation becomes P = (V/R)V, which simplifies to P = V²/R. This indicates that voltage squared divided by resistance provides another expression for power.

Thus, all three formulas are valid expressions for calculating power in a resistor, reinforcing the idea that multiple relationships exist to define how power, voltage, current