A 15.0 kΩ resistor is hooked up to a 45.0 V battery in a circuit with a switch.
a.) What is the current flowing through the resistor?
b.) What is the power dissipated by the resistor?
V=IR
I=V/R=45/15000=3*10-3 A.
Power = VI joules
Power = VI watts
To solve this problem, you will need to use Ohm's Law and the formula for power.
a.) What is the current flowing through the resistor?
According to Ohm's Law, the current (I) flowing through a resistor can be calculated using the formula:
I = V / R
Where:
I = Current (in Amperes)
V = Voltage (in Volts)
R = Resistance (in Ohms)
In this case, we are given a voltage of 45.0 V and a resistance of 15.0 kΩ. However, it's important to note that Ohm's Law applies to resistance in Ohms, so we need to convert the kilo-ohm value to ohms:
15.0 kΩ = 15.0 × 1000 Ω = 15,000 Ω
Now we can substitute these values into the formula:
I = 45.0 V / 15,000 Ω
I ≈ 0.003 A (rounded to three decimal places)
Therefore, the current flowing through the resistor is approximately 0.003 Amperes.
b.) What is the power dissipated by the resistor?
The power dissipated by a resistor can be calculated using the formula:
P = I^2 * R
Where:
P = Power (in Watts)
I = Current (in Amperes)
R = Resistance (in Ohms)
Using the current value we calculated in part a (I ≈ 0.003 A) and the resistance value given (15.0 kΩ = 15,000 Ω), we can substitute these values into the formula:
P = (0.003 A)^2 * 15,000 Ω
P ≈ 0.045 W (rounded to three decimal places)
Therefore, the power dissipated by the resistor is approximately 0.045 Watts.