The Amateur Radio Project - Resistors in series and parallel; power The Amateur Radio Project - Resistors in series and parallel; power

# Resistors in series and parallel; power

### Explanation of question 14

A resistor is connected across a 12V supply. A 6V supply is now substituted for this supply. The new power will be

1. four times the original power
2. twice the original power
3. one-half the original power
4. one-quarter the original power

   The answer is :- 4

   Assume the resistor as a value of 24 Ohms. I=V/R

   At 12v I would be I = 12/24 = 0.5Amps. W = V x I or W= 12 x 0.5 = 6w

   At 6v I would be I=6/24 = 0.25Amps. W = V x I or W = 6 x 0.25 =1.5w

   1.5w is one-quarter of the original power of 6w

   OR:-

   P = V2 / R.

   R is a constant, therefore, P V2. <

   New power will be (½)2 = ¼ of the original power.

R is a constant, therefore, P V2. <

New power will be (½)2 = ¼ of the original power.

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