Question

two identical resistors each of resistance 12 ohm are connected(i) in series (ii) parallel in turn to a battery of 6 volt calculate(a) power dissipated in each combination during one second(b) the ratio of power consumed in series to parallel combination.​

Answer 1

(1) IN SERIES

Total Resistance in Series = R₁ + R₂

                                         = 12 + 12

                                         = 24 Ohm

Potential Difference(V) = 6V

Current(I)= V/R

              =6/4

              =3/2Ampere = 1.5A

Power = VI

          = 6*1.5

          =9 Watts

(2) IN PARALLEL

1/R₁ + 1/R₂ = 1/Rₐ (Where Rₐ is the total resistance)

1/2 + 1/2 = 1/Rₐ                                            

2/2 =1/Rₐ

1/1 = 1/Rₐ      

1 Ohm = Rₐ

Voltage (V)= 6Volts

Current(I)= V/R

              = 6 Ampere

Power= VI

         = 6*6

         = 36 Watt

Power in series : Power in Parallel

1.5 : 36

1 :24

Answer 2

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