Electronics - Parallel Circuits - Discussion

21. 

When a parallel 5 k resistor and a 25 k resistor have a 10 V supply, what is the total power loss?

[A]. 2.4 mW
[B]. 3.3 mW
[C]. 24 mW
[D]. 33 mW

Answer: Option C

Explanation:

No answer description available for this question.

Sangeetha said: (Jan 6, 2011)  
RT=4.166k
V=10
P=V2/RT
P=24mW

Jayasree said: (Feb 2, 2011)  
How can RT be 4.166 k ?

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

Mohammad Umair Kabir said: (Jul 26, 2011)  
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: (Jan 27, 2012)  
Since, r1||r2

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

Abi said: (Jun 2, 2012)  
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: (Jun 29, 2012)  
Rt=25*5/30.

=5/6 kohm.

P=V*V/R.

=100*6/5.

=24mW.

Shahid said: (Sep 25, 2012)  
@Jayashree.

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

Anita Nagar said: (Jul 5, 2013)  
RT = R1*R2\(R1+R2) = 4.166.

P = V^2\R = 24.03MW.

Prashant said: (Jun 4, 2017)  
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.

Mani said: (May 1, 2018)  
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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