Electronics - Parallel Circuits - Discussion

Discussion Forum : Parallel Circuits - General Questions (Q.No. 19)
19.
What is the total power of the circuit?

mcq4_1009_1.gif
2.2 W
4.2 W
6.2 W
8.2 W
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
15 comments Page 1 of 2.

Himanshu Verma said:   1 decade ago
Equivalent resistance of the circuit R = R1||R2||R3
R = 1K||2K||3K = 6/11 K

Power P = V2/R = (48*18*11)/6K = 4.224 mA

Naseem Ul aziz said:   1 decade ago
p1=v^2/r=2.304 w
p2=1.152 w
p3=0.768 w
pt=4.224 w

Bharti S.K. said:   1 decade ago
In the given Ckt. Eq. resistance of the circuit Req. = R1||R2||R3
1/R = 1/1K+1/2K+1/3K = 11/6 K
So R = 6/11 K

Hence Total Power P = V.V/R = (48*48)11/6K
= 4.224 mA

Snehashis said:   1 decade ago
In the given circuit eq resistance=R1//R2//R3 = 6/11k
now total current flowing in the circuit isi=v/r=48/(6/11)*1000

=0.088A

Now power loss p=i*iR=4.2w

Nehha said:   1 decade ago
The resistances are connecting in parallel. so
1/R=1/1+1/2+1/3=11/6
Or R=6/11
Power=V.V/R=48.48.11/6
P=4.22w

Divya vinoth said:   1 decade ago
Rt=1kohms( by parallel resistance formula)
P=V^2/R
P=48^2/1k
P=4.224W

Ali Umar said:   1 decade ago
Can any body give give justification.

In Parallel connection Formula is 1/R = 1/R1 + 1/R2 + 1/R3.

So 1/R = 1/1 + 1/2 + 1/3 = 11/6 ==> are = 6/11 kOhm Answer.

I another way are = R1*R2*R3/ (R1+R2+R3) =1*2*3/ (1+2+3) =1kOhm Answer.

My question is Why my answer is coming different. Please help me.

Rathi said:   1 decade ago
@Ali Umar.

Another way are = R1*R2*R3/(R2*R3 + R1*R3 + R1*R2).

1*2*3/(2*3 + 1*3 + 1*2) = 6/11.

SAN said:   1 decade ago
I = V/R THEN TO FIND POWER P = V*I.

I1 = V1/R1 = 48/1000 = 0.048.
I2 = V1/R2 = 48/2000 = 0.024.
I3 = V1/R3 = 48/3000 = 0.016.

TOTAL CURRENT = I1+I2+I3 = 0.048+0.024+0.016 = 0.088.

POWER = V*I = 48*0.088 = 4.224WATTS.

Alka Sutar said:   1 decade ago
Equation of resistance of circuit = R = R1||R2||R3.

R = 6/11K.
POWER = V2/R.
= (48*48)*11/6K.
= 4.224 mA.


Post your comments here:

Your comments will be displayed after verification.