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Posted by on Thursday, August 23, 2012 at 12:34am.

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 - , Thursday, August 23, 2012 at 8:48am

    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 - , Thursday, August 23, 2012 at 7:33pm

    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 - , Saturday, August 25, 2012 at 2:42am

    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 - , Monday, May 20, 2013 at 4:48am

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