# PHYSICS

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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 value of R should be chosen in order to supply each bulb with a voltage of 2.4 V ?For what value of R would the potential difference across each of the bulbs be 2.4 V ?

• PHYSICS -

The series resistor R should be chosen so as to have a voltage drop of 7.6V(10-2.4)across it.
The currents in the two bulbs:
i1=2.4/3.0 = 0.8A
i2=2.4/2.5 = 0.96A

Total current in the circuit = i1+i2
= 1.76A
R = 7.6V/1.76A
= 4.3 Ohms

• PHYSICS -

How do i solve this part:
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?

• PHYSICS -

The current through each bulb is the voltage across it divided by its resistance. In this case the bulbs are in parallel and voltage across them is 2.4V.

• PHYSICS -

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