Question

Two resistors a and b of resistance 4 ohm and 6 ohm respectively are connected series.the combination is connected across a 6 volt battery of negligible internal resistance .calculate:
(i)the power supplied by the battery
(ii) the power dissipated in each resistor.

Answer 1

Answer:V = 6 , R1 = 4 , R2 = 6

Total Resistance

1/R = 1/4 + 1/6

1/R = 3/12 + 2/12

1/R = 5/12

R = 12/5 = 2.4 ohms

Power supplied by battery

P = V^2/R

P = (6*6)/2.4 = 36/2.4 = 15 Watt

Power dissipated in R1

P = V^2/R

P = (6*6)/4 = 36/4 = 9 Watt

[ P.d is same across every resistor in Parallel combination ]

Power disspitated in R2

P = 36/6 = 6 Watt

Explanation:

Answer 2

Explanation:

i)P=V^2/R=36/10=3.6 watt

ii) From Ohm's law,

I=V/R

or,I=6/10=0.6 A

In series connection of resistances, total current across each of the resistors remains constant

Power dissipated in 4 ohm resistor=I^2R=(0.6)^2x4=1.44 Watt

Power dissipated in 6 ohm resistor=I^2R=(0.6)^2x6=2.16 watt

(This is the correct answer because the resistances are connected in series)