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

Two bulbs are connected in parallel across a source of emf E = 12.0 V with a negligible internal resistance. One bulb has a resistance of 3.0 Ω , and the other is 2.5 Ω . 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 ?

Part C

For what value of R would the potential difference across each of the bulbs be 2.4 V ?

Express your answer in ohms using three significant figures.

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R =
Ω

ough the bulb of resistance 3.0 Ω, and Ib2 the current through the bulb of resistance 2.0 Ω .

Enter each current value in amperes using three significant figures and separated by commas.

Ib1, Ib2 =


A,A

Respuesta :

Answer:

Explanation:

The concept of Ohm's law is applied, from V = IR

Part C )

  • Considering the current passing through resistor R1;
  • Ib1 = V/R1 = 2.4V/3Ω = 0.8A

similarly, the current passing through resistor R2 ;

  • Ib2 = V/R2 = 2.4V/2.5Ω = 0.96A
  • Hence, total current = Ib1 + Ib2 = 0.8A + 0.96A = 1.76A

from E = V + IR

  • R = E - V /I = 12V - 2.4V /1.76A
  • R = 5.45Ω

Current passing through the two bulbs = 0.8A and 0.96A

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