The Calculation of Resistance and Power in a Light Bulb Circuit
Understanding Resistance and Power in a Light Bulb Circuit
We have a circuit with a battery and a light bulb. A voltage of 4.50 V is applied across the light bulb. This causes a current of 120.00 mA to pass through the filament in the light bulb.
a. Calculate the resistance R of the light bulb in W
The resistance of the light bulb is 37.5 ohms, and the power dissipated in the light bulb is 0.54 W.
To calculate the resistance R of the light bulb, we can use Ohm's Law, which states that resistance is equal to voltage divided by current:
R = V / I
In this case, V is 4.50 V and I is 120.00 mA (or 0.120 A, since 1 mA is equal to 0.001 A).
So, R = 4.50 V / 0.120 A = 37.5 ohms
b. Calculate the power P dissipated in the light bulb in W
To calculate the power P dissipated in the light bulb, we can use the formula:
P = V * I
Again, V is 4.50 V and I is 120.00 mA (or 0.120 A).
So, P = 4.50 V * 0.120 A = 0.54 W
In summary, the resistance of the light bulb is 37.5 ohms, and the power dissipated in the light bulb is 0.54 W.
We have calculated the resistance and power in the light bulb circuit. Can you explain the importance of understanding these values in electrical circuits? Hi! I'll be happy to help you with your question. Please note that I will provide a concise answer as per my guidelines.a. To calculate the resistance (R) of the light bulb, we can use Ohm's Law:
V = I * R Where V is voltage (4.50 V), I is current (120.00 mA or 0.120 A), and R is resistance. R = V / I R = 4.50 V / 0.120 A R = 37.5 ohms
b. To calculate the power (P) dissipated in the light bulb, we can use the formula: P = V * I Where P is power in watts (W), V is voltage (4.50 V), and I is current (0.120 A). P = 4.50 V * 0.120 A P = 0.54 W To know more about ohms visit: