Electronics - Parallel Circuits - Discussion

Discussion :: 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?

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

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.