A torch bulb is rated 2.5V and 750mA. Two identical resistors, each of resistance 2 ohm , are connected in series and parallel to a battery of 12 V Calculate ratio of power consumed in two cases.

Open in App

Solution

IN SERIES

Total Resistance in Series = R₁ + R₂
= 2 + 2
= 4 Ohm

Potential Difference(V) = 12V
Current(I)= V/R
=12/4
=3Ampere
Power = VI
= 123
=36 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)= 12Volts
Current(I)= V/R
= 12 Ampere
Power= VI
= 1212
= 144 Watt

Power in series : Power in Parallel
36 : 144
1 : 4

Regards

Suggest Corrections

Similar questions

Q. Two heating coils of resistances

10 Ω

and

20 Ω

are connected in parallel and connected to a battery of emf

12 V

and internal resistance

1 Ω

. The power consumed by them are in the ratio

Q. Three identical cells each of emf 2V and internal resistance 1

Ω

are connected in series to form a battery. The battery is then connected to a parallel combination of two identical resistors, each of resistance 6

Ω

. Find the current delivered by the battery.

Two resistors of $2.0 Ω$ and $3.0 Ω$ are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case, draw a circuit diagram and calculate the current through the battery.

Q. Two resistors of

2.0 Ω

and

3.0 Ω

are connected (a) in series, (b) in parallel, with battery of 6.0 V and negligible internal resistance. For each case

Join BYJU'S Learning Program

​​A torch bulb is rated 2.5V and 750mA. Two identical resistors, each of resistance 2 ohm , are connected in series and parallel to a battery of 12 V Calculate ratio of power consumed in two cases.

A torch bulb is rated 2.5V and 750mA. Two identical resistors, each of resistance 2 ohm , are connected in series and parallel to a battery of 12 V Calculate ratio of power consumed in two cases.