# Electronics - Parallel Circuits - Discussion

### Discussion :: Parallel Circuits - General Questions (Q.No.4)

4.

What is the total power loss if 2 k and 1 k parallel-connected resistors have an IT of 3 mA?

 [A]. 6 W [B]. 36 W [C]. 6 mW [D]. 36 mW

Explanation:

No answer description available for this question.

 Rama said: (Oct 31, 2010) We know that, P=v*i v=i*R so P=i2*R 1/R=(1/2)+(1/1) R=2/3kohm=(2*10^3)/3ohm and i=3mA=3*10^-3A p=(3*10^-3)^2*(2*?10^3/3) P=6*10^-3W P=6mW

 Mamatha.M said: (Feb 14, 2011) WKT, P=I2*R where Rp=R1R2/R1+R2 Rp=(2*1)/(2+1) Rp=2/3k ohm P=(3m)^2*(2/3)k P=6mW

 Sree said: (Feb 17, 2011) The formula for resistor when 2 resistors are connected in parallel is 1/R=1/R1+1/R2.

 Gautam said: (Mar 1, 2011) Here r=2000/3 & i^2=(3*10^-3)^2=9*10^-6 we know P=i^2*r=6*10^-3=6mW

 Ramprabu said: (Mar 26, 2011) P=I^2(R) R=>2K AND 1K IN PARALLEL HENCE EQUALENT R VALUE IS (2/3) P=(3m)^2(2/3) =>(9u)(2/3) =>6uw SO I CONCLUDE HERE OPTION A IS CORRECT

 Syed said: (Apr 10, 2011) I agree with ramprabhu answer. Opion "A" is correct.

 Devaki said: (Apr 28, 2011) P=I^2(R) R= 2K PARELLEL TO 1K OHM = 2/3 K OHM P= 3mA * 3mA * 2/3 * 10^3 = 6mW SO I CONCLUDE THAT OPTION 'C' IS CORRECT

 Naseem Ul Aziz said: (May 6, 2011) p=vi p1=? p2=? pt=? p1=???? vt=ir rt=666.666 ohm vt=2 volt p1=2m w p2=4mw pt=p1+p2=6 m w

 Chandu said: (May 16, 2011) PARALLEL CONNECTION:- R eq = [(2 X 1) (10^3 X 10^3)]/[(2+1) 10^3] 2/3 10^3 OHMS.......... P= I * I * R eq (3*3*2/3)(10^-3)(10^-3)(10^3) 6mw ans

 Ramakant said: (Jul 7, 2011) Rt=2/3k ohms where we know P=I2*R then P=(3*10^-3)(3*10^-3)(2/3*10^3) P=6*10^-3w P=6mw THEREFORE 'C' IS THE CORRECT ANSWER

 Gopi said: (Sep 22, 2011) P=I^2*R R=(2/3)*10^-3 I=3*10^-3 P=(3*10^-3)(3*10^-3)(2/3)*10^-3 P=6w

 Nandini said: (Oct 30, 2011) Yes, I also agree vit rambabu.

 Sunil Kumar said: (Nov 8, 2011) Parallel Connection Req = R1*R2/R1+R2. Req = 2000*1000/2000+1000 = (2000/3000) ohms i.e (2/3)K ohms. P = I^2 * Req. = (3*10^-3)^2 * (2/3) * 10^3 = 9 * 10^-6 * (2/3) * 10^3 = 6 m Watts

 Vinay said: (Nov 16, 2011) Current will divide as 2mA in 1K and 1mA in 2k resistors so power is given by p=i^2*R power at 2k=1^2*2 =2mw |||'y at 1k=2^2*1 =4mw Total power=2mw+4mw =6mw

 Bk Rout said: (Nov 20, 2011) Vinays answer is right. So option C is correct.

 Piyush Suthar said: (Dec 14, 2011) Thats simple 1/r=1/2+1/1 r=2/3 p=i^2*r =(3ma)^2*2/3 =9ma*2/3 =6mw

 Linton said: (Jan 13, 2012) @piyush suthar How 1/2 + 1/1 will makes 2/3 ? 1/2 = (.5)+(1) = 1.5 = 3/2 not 2/3.

 Abhishek Gupta said: (Feb 5, 2012) 1*2=1/2*3=2*3=6W

 Jonathan said: (Apr 8, 2012) Ramprabu is wrong, he should include 10^3 in total resistance, you'll get ((2/3)x10^3)(3mA)^2 = 6mW

 Anjali.R said: (Apr 10, 2012) P=i^2*r P=3^2*(2/3) P=6mw

 Sudhakar said: (Jun 15, 2012) Here total current given is It=3mamp, And R1=2kohm, R2=1kohm are in parallel, So equivalent resistance of 2 parallel resistors is Req=R1*R2/ (R1+R2). So Req=2kohm*1kohm/ (2kohm+1kohm). = (2*1*kohm*kohm) / (3kohm). =2kohm/3. Now total power loss in a circuit is=It^2 (Req). = (3mamp*3mamp) (2kohm/3) =6mw is the correct answer.

 U.Suruthi said: (Jul 30, 2012) R = (2*1)/(2+1) = 2/3. I = 3*10^-3. P = I^2*R. P = (3*10^-3)^2*(2/3). P = 6*10^-3w.

 Prasanna said: (Aug 19, 2012) Ramprabhu is wrong, Resistance value here given is Kohms and he consider it to be only ohms

 Abc said: (Oct 31, 2012) What would be the total power loss had the circuit been connected in series?

 Vinod said: (Dec 5, 2013) parallel resistors: 1/r = 1/r1+1/r2. 1/r = 1/2+1/1. 1/r = 3/2. r = 2/3. p = v*i. v = i*r. p = i^2*r. p = 3^2*2/3. p = 9*2/3. p = 6mA.

 Hims said: (Mar 1, 2014) Rp = 2*1/2+1 = 2/3 kohm = (2/3) * 10^3 ohm. P = i2*r. i = 3mA = 3*10^-3A. P = (3*10^-3)A*(3*10^-3)A* (2/3)* 10^3 ohm. P = (3*3*2/3)W * (10^-3*10^-3*10^3). P = (6)W*(10^-3) = 6*10^-3W. P = 6mW.

 Irshad K A said: (May 3, 2014) R = 2/2+1/2. R = 3/2. R = 2/3. R = 0.6668. P = I2R. P = 3*3*0.6668. P = 6(approximately).

 Asha said: (Oct 11, 2014) P = VI.....1. V = IR......2. Substitute above equation in 1, we get P = I^2*R. R = 2*1/2+1, R = 2/3, P = (3*10^-3)^2*(2/3)*10^3 ohm, p = 6mW.

 Azhagusurya said: (Feb 20, 2015) Power loss p = i^2/Rt. Rt = (r1*r2)/(r1+r2). Rt = (2*1)M/(2+1)k. Rt = (2/3)k. p = (3^2)u*(2/3)k. p = (18/3)m= 6mW.

 Rashmi said: (Jun 24, 2015) V = IR BY OHM'S LAW. From the following data: I = 3 m. R1 = 1k. R2 = 2k. When resistors are connected in parallel. Req is given by Req = R1*R2/R1+R2. P = I^2*Req. P = (3m)^2*(2/3)k. P = 6 mW.

 Ana said: (Nov 14, 2015) 1/R = 1/2+1/1. 1/R = 1+2/2. R = 2/3 Kohm. V = i*R. V = 3 mA*2/3 Kohm. V = 2 v. P = V*I. P = 2v*3 mA. P = 6 mW.

 Addy said: (Jul 29, 2016) Simply, Low resistance = Higher current flow. So, we get applied voltage by dividing current in the ratio 2k/1k = 3mA. 1k = 2mA and 2k = 1mA. Hence, applied voltage is 2V. Now, we have I = 3mA, V = 2v. P = V * I. = 2 * 3mA. = 6mA. ANSWER 'C' IS CORRECT.

 Chandrashekhar Manekar said: (Aug 25, 2016) Actually power( P) = I^2(R). By using this, we can solve the problem.

 Krish said: (Sep 10, 2016) Two parallel resistor so equivalent resistor is 2/3 and also current 3 mA. So power loss is (p = Isquare R) 9 * 2/3 = 6 mA.

 Divya said: (Dec 13, 2016) P = VI -----> 1. V = IR -----> 2. Substitute above equation in 1, We get P = I^2 * R. R = 2*1/2 + 1, R = 2/3, P = (3*10^-3)^2*(2/3)*10^3 ohm, p = 6mW.

 Pijush Patra said: (Jul 7, 2017) What is the total resistance of 5 kilo ohm, 5mege ohm connected in parallel?

 Shah Qaisar said: (Nov 10, 2018) P = i^2R. I = 3mA= 0.003A, R = 2000 * 1000/2000+1000= 666.6 ohm, P = (0.003 * 0.003)* 666.6= 0.00599~6mW. P =(.003 * .003). Option C is correct.

 Sameer said: (Jan 8, 2019) Power=Voltage * Current. Voltage= Current * Resistance, Resistors in parallel (1/R=1/R1+1/R2) OR (R=R1*R2/R1+R2), so, 1/R= 1/2+1/1 = 3/2KOhm , R= 2/3 KOhm, P=V*I = (I*R)* I. P= I*I*R, P=3*3*2/3, P=9*2/3, P=18/3, P=6mW.

 Moksha said: (Feb 1, 2019) p = I^2R. 1/R=(1+2)/(1*2) = 3/2, R = 2/3k ohms, p = (3m)^2 * 2/3k = 6mw.