To practice Problem-Solving Strategy 26.1 Resistors in Series and Parallel. Two

Question

To practice Problem-Solving Strategy 26.1 Resistors in Series and Parallel. Two bulbs are connected in parallel across a source of emf epsilon = 8.00 V with a negligible internal resistance. One bulb has a resistance of 3.0 Ohm, and the other is 3.0 Ohm. 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? The bulbs in this circuit can be considered resistors. So if you were to draw the resistor network, you could replace the bulbs with two resistors of resistance R_1 = 3.0 Ohm and R_2 = 3.0 Ohm. Here is an example of what your drawing may look like: Which of the following statements are correct for the circuit described in this problem?

Explanation / Answer

The equivelent of parallel combination of resistors:

R = R1R2 / ( R1 + R2 ) = 3•3 / 6 = 1.5

As 2.4 volts across them, the current is I = V /R = 2.4/1.5 =1.6 amps.
and the unknown R = (9 – 2.4) / 1.6 = 4.12

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The current through the 3 is 2.4 V / 3 = 0.8 A

and current through another resistor is same, that is

The current through the 3 is 2.4/3 = 0.8 amps

and they add up to 1.6 amps, which they should.

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As they are connected in parallel, the voltage across each bulb is same, regardless of the resisistances.

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