what is the power of a 6 ohm bulb when 3 of such bulbs ar econnected in a 12 volt parallel circuit?
power= 3*V^2/R=3*144/6 watts
To find the power of a 6-ohm bulb connected in parallel with two other identical bulbs in a 12-volt circuit, we can use the formula for power:
P = V^2 / R
Where:
P = Power
V = Voltage
R = Resistance
In this case, the voltage is 12 volts and the resistance of each bulb is 6 ohms. Since the bulbs are connected in parallel, the total resistance is calculated as follows:
1 / R_total = 1 / R1 + 1 / R2 + 1 / R3
R_total = 1 / (1/R1 + 1/R2 + 1/R3)
Substituting the values:
R_total = 1 / (1/6 + 1/6 + 1/6) = 2 ohms
Now we can calculate the power:
P = V^2 / R_total = 12^2 / 2 = 144 / 2 = 72 watts
So, the power of each 6-ohm bulb connected in parallel with two others in a 12-volt circuit is 72 watts.
To calculate the power of a bulb, you need to use the formula: power (P) = voltage (V) squared divided by resistance (R).
In this case, you have three 6 ohm bulbs connected in a parallel circuit, which means the total resistance (Rtotal) is given by the reciprocal of the sum of the reciprocals of the individual resistances. So, for three 6 ohm bulbs, the total resistance (Rtotal) would be:
1/Rtotal = (1/6) + (1/6) + (1/6) = 3/6 = 1/2
Therefore, Rtotal = 2 ohms.
The voltage (V) in the circuit is given as 12 volts.
Now, you can calculate the power (P) of the bulbs:
P = V^2 / Rtotal
P = 12^2 / 2
P = 144 / 2
P = 72 watts
So, the power of each bulb in a parallel circuit with three 6 ohm bulbs would be 72 watts.