Question

Two identical resistors each of resistance 50 ohm are connected in series, in parallel in turn to a battery of 10v calculate the ratio of power consumed in the combination of resistance in the two cases

Answer 1

Answer:

1)in series

10 + 10 = 20Ω

now find the current

I=V/R

 =6/20

 =.3A

now power

P1=V(I)

  =6×.3

 = 1.8W

2)in parallel

1/R=1/R+1/R

 =1/10+1/10

R  =5Ω

now current

I=V/R

 =6/5

 =1.2A

power

P2=V×I

 =6×1.2

 =7.2W

Ratio

P1/P2=1.8/7.2

         =.25W

Explanation:

Answer 2

Answer:

The ratio of powers in series and the parallel combination will be 1:4.

Explanation:

case I:

when resistors in the circuit are connected in series

the equivalent resistance will be=R1 + R2=50Ω + 50Ω=100Ω

power in series combination(Ps) will be $V^{2} /Rs$

now let's put the values in the given equation,

Ps=$(10)^{2} /100$=1 W      

case II:

when the resistors in the circuit  are connected in parallel

the equivalent resistance will be(Rp),

1/Rp=1/R1 +1/R2

Rp=25Ω

power in parallel combination(Pp) will be $V^{2} /Rp$

lets put the values in the given equation

Pp=$(10)^{2} /25$=4 W

the ratio of powers in series and the parallel combination will be,

Ps/Pp=1/4

i.e.

Ps:Pp=1:4