Question

In the network shown in the figure, power dissipated in 3ohm is 12W power dissipated in 4 ohm will be?

Answer 1

Answer:

4W

Explanation:

from 3 ohm resistor

P = v^2/r

12 = v^2/3

v = 6

now in 4 ohm resistor

v = ir

6 = i x 6

i = 1amp

P = i^2 x R

   = 1 x 4

   = 4W

Answer 2

Answer:

4W

Explanation:

According to the question,

We know that,

$P = VI =V^2/R = I^2R\\\\\\V = IR$

from 3 ohm resistor

$P = V^2/R\\\\12 = V^2/3\\\\V = 6$

Now in a 4-ohm resistor

total resistance is equal to4ohm side is  6 ohm

total current flow in a 4-ohm resistor is-

$V = IR\\\\6 = I X 6\\\\I = 1amp$

then,

the power dissipated in 4-ohm will be-

$P = I^2 R\\\\= 1 X 4\\\\= 4W$

Hence, the power dissipated in 4-ohm will be 4W.