# Electronics - Parallel Circuits - Discussion

Discussion Forum : Parallel Circuits - General Questions (Q.No. 21)
21.
When a parallel 5 k resistor and a 25 k resistor have a 10 V supply, what is the total power loss?
2.4 mW
3.3 mW
24 mW
33 mW
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Santhanakumari said:   2 years ago
Please clear the solution in detail. @Deepika.

p= 24 * 10^-3.
And P = 24 mwatts.

Santhanakumari said:   2 years ago
Please clear the solution in detail. @Deepika.

p= 24 * 10^-3.
And P = 24 mwatts.

Deepika.S said:   4 years ago
P=v*v/R.

R= (R1*R2) / (R1+R2).

R= (5*25) / (5+25).

R=4.16*10^3ohms.

P= (10*10) / (4.16*10^3).

P=24*10^-3.

P=24mwatts.

Mani said:   5 years ago
We know p=v*v/R.

Given R1&R2 are connected in parallel,
R1 = 25k&R2=5k,
R = R1llR2,
R = R1*R2/R1+R2=4.16kohms,
P = 10*10/4.16*10^3,
P = 24mwatts.

Prashant said:   6 years ago
R1 =5K THATS WHY P1 =10*2/5K =20mw (P=V2/R).
R2= 25K THATS WHY P2 = 10*2/25K =4mw.(P=V2/R).

SO,
P = P1+P2 =24mw.

ANITA NAGAR said:   1 decade ago
RT = R1*R2\(R1+R2) = 4.166.

P = V^2\R = 24.03MW.

@Jayashree.

Both the resistors are parallel to each other means (25*5) / (25+5) =4. 16.

Rt=25*5/30.

=5/6 kohm.

P=V*V/R.

=100*6/5.

=24mW.

P=VI
BUT I=V/R
SO P=V.V/R
R=R1.R2/R1+R2
BCOZ RESISTORS ARE PARALLEL HERE.
R=4.166K ohms.
P=100/4.166K
=24mw.