Physics, asked by reehu786, 11 months ago

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.

Answers

Answered by rohitrajesh54

(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

kunal111128: wrong

JRJosh: you did a mistake that in parallel the resistance in each resistor is 12 ohm

reehu786: the answer should be 1:4 I think

kunal111128: current =V square /r

Answered by JRJosh

I think this answer is correct

Attachments:

kunal111128: wrong

kunal111128: you have find heat instead of power

JRJosh: y??

reehu786: thanks

JRJosh: I find both heat and power. The question is both heat and power. You won't see answer

reehu786: no. the question is asking about only power.

JRJosh: a) power dissipated in one second it is heat. b) power consumed

kunal111128: your mathod is correct but the heat=I square ×r×t

JRJosh: yeah i know but i did it wrong

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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.​

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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.