Two bulbs are connected in parallel across a source of emf EMF = 10.0V with a negligible internal resistance. One bulb has a resistance of 3.0 Omega , and the other is 2.5 Omega . A resistor R is connected in the circuit in series with the two bulbs.
What is the current through each individual bulb? Let I1 be the current through the bulb of resistance 3.0 Omega and I2 the current through the bulb of resistance 2.5 Omega?
2.4/3=.8 A
2.4/2.5=.96 A
To find the current through each individual bulb, we first need to find the total resistance of the circuit.
In a parallel circuit, the reciprocal of the total resistance is equal to the sum of the reciprocals of the individual resistances. So, we can calculate the total resistance (R_total) as follows:
1 / R_total = 1 / R1 + 1 / R2
Given that R1 = 3.0 Ω and R2 = 2.5 Ω, we can substitute these values into the equation:
1 / R_total = 1 / 3.0 + 1 / 2.5
To add the fractions, we need a common denominator. The common denominator is 7.5, so we can rewrite the equation as:
1 / R_total = (2.5 + 3.0) / 7.5
Simplifying the numerator, we get:
1 / R_total = 5.5 / 7.5
To isolate R_total, we take the reciprocal of both sides of the equation:
R_total = 7.5 / 5.5
Now we can calculate the value of R_total:
R_total = 1.364 Ω (rounded to three decimal places)
Next, we need to find the total current (I_total) flowing through the circuit. We can use Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R):
I_total = EMF / R_total
Given that EMF = 10.0 V, we can substitute this value into the equation:
I_total = 10.0 / 1.364
Calculating this, we get:
I_total ≈ 7.323 A (rounded to three decimal places)
Since the bulbs are connected in parallel, the current flowing through each bulb will be the same as the total current (I_total).
Therefore, I1 = I_total = 7.323 A and I2 = I_total = 7.323 A.
So, the current through the bulb of resistance 3.0 Ω (I1) is approximately 7.323 A, and the current through the bulb of resistance 2.5 Ω (I2) is also approximately 7.323 A.