In the Fig above, Current I passes through the combination, if the power dissipated in the 5 ohm resistor is 40W, then the power dissipated in the 10 ohm resistor is

Mhoyinholuwar

5 years ago

Since in a parallel connection there is different current but same voltage in the circuit.
Where P1= V²/R
40 = V²/5
Then V²=200v
Since same voltage is applied
P = 200/10
P = 20W

christy1234

9 years ago

pls explain

De sage

1 year ago

p1R1= P2R2
P1=40w
R1=5
P2=?
R2=10
P2=40×5/10
P2=200/10
P2= 20w

Overwatch886

2 years ago

Power Dissipated=I
2
×R

Given that the current
�
I in both resistors are different, but from
�
=
�
�
V=IR, we have
�
=
�
�
I=
R
V
​
.

Substituting
�
=
�
�
I=
R
V
​
into the previous formula, we get:

Power Dissipated
=
(
�
2
×
�
�
2
)
Power Dissipated=(
R
2

V
2
×R
​
)

Power Dissipated
=
�
2
�
Power Dissipated=
R
V
2

​


Given that
40
=
�
2
5
40=
5
V
2

​
in the
5
Ω
5Ω resistor, we can solve for
�
2
V
2
:

�
2
=
40
×
5
=
200
V
2
=40×5=200

�
=
200
V=
200
​


Then, the power dissipated in the
10
Ω
10Ω resistor is:

Power Dissipated
=
�
2
�
=
(
200
)
2
10
=
200
10
=
20
watts
Power Dissipated=
R
V
2

​
=
10
(
200
​
)
2

​
=
10
200
​
=20 watts

So, option B is correct.

Efranklyn

4 years ago

The truth is that the answer is 20W well detailed by Moyin

Olasunkanmi2004

5 years ago

is answer not 80w

Overwatch886

2 years ago

Power Dissipated = I^2 x R
but the current, I in the both resistors are different but from V=IR
I = V/R
Then put in the previous formula, we should have (V^2 x R) / R^2
then 40 = V^2/5
V^2 = 200
V = √200
For the 10 ohm resistor, the power dissipated is V^2/R
= ((√200)^2)/10
= 200/10
= 20 watts
That is option B

olyezema

4 years ago

The correct answer is 80watts
P=I²R
40=I²5
40/5=√8
P=I²R
P=8×10
P=80watts