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
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Sangeetha said:   1 decade ago
RT=4.166k
V=10
P=V2/RT
P=24mW

Jayasree said:   1 decade ago
How can RT be 4.166 k ?

Since both resistors are parallel to each other RT=2.5 ohms na ?

Mohammad Umair Kabir said:   1 decade ago
Since resistors are parallel so RT will be

1/RT = 1/R1 + 1/R2
1/RT = 1/5 + 1/25 = 6/25
RT = 25/6 or 4.16K

I = V/R = 10/(25/6) = 2.4 mA

So P = VI = (10)(2.4)= 24 mW

Chinnu said:   1 decade ago
Since, r1||r2

rt = 4.166
P = V*V/R
=> P=24.03mW

Abi said:   1 decade ago
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.

Viju said:   1 decade ago
Rt=25*5/30.

=5/6 kohm.

P=V*V/R.

=100*6/5.

=24mW.

Shahid said:   1 decade ago
@Jayashree.

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

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

P = V^2\R = 24.03MW.

Prashant said:   8 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.
(1)

Mani said:   7 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.


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