In a series circuit there is 12V running with 3 bulbs with the following: 1Ω, 2Ω, and 3Ω.
Calculate the potential difference across each light bulb. Assume r = 0Ω.
current in circuit=12/(1+2+3)=2a
potential across each bulb= I^2 R = 4R
6Ω total resistance , so 2A current
potential difference = current * resistance
I = E/Rt = 12/(1+2+3) = 2A.
V1 = I * R1 = 2 * 1 = 2 Volts.
V2 = I * R2 = 2 * 2 = 4 Volts.
V3 = I * R3 = 2 * 3 = 6 Volts.
To calculate the potential difference across each light bulb in a series circuit, we need to use Ohm's Law, which states that V = I * R, where V represents the potential difference (voltage) across the circuit component, I represents the current flowing through the component, and R represents the resistance of the component.
In this case, we have a series circuit with a total voltage of 12V and three bulbs with resistances of 1Ω, 2Ω, and 3Ω. Since the bulbs are connected in series, the current flowing through each bulb will be the same.
To find the current flowing through the circuit, we can use Ohm's Law again. The total resistance in a series circuit is the sum of the individual resistances. Therefore, the total resistance of the circuit is 1Ω + 2Ω + 3Ω = 6Ω.
Using Ohm's Law, we can find the current flowing through the circuit:
V = I * R
12V = I * 6Ω
Rearranging the equation to solve for I:
I = V / R
I = 12V / 6Ω
I = 2A
Now that we know the current flowing through the circuit is 2A, we can calculate the potential difference across each bulb using Ohm's Law:
V1 = I * R1
V1 = 2A * 1Ω
V1 = 2V
V2 = I * R2
V2 = 2A * 2Ω
V2 = 4V
V3 = I * R3
V3 = 2A * 3Ω
V3 = 6V
Therefore, the potential difference across each light bulb is:
V1 = 2V
V2 = 4V
V3 = 6V