Electronics - Parallel Circuits - Discussion
Discussion Forum : Parallel Circuits - General Questions (Q.No. 19)
19.
What is the total power of the circuit?
Discussion:
15 comments Page 2 of 2.
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
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
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
now total current flowing in the circuit isi=v/r=48/(6/11)*1000
=0.088A
Now power loss p=i*iR=4.2w
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
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
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
p2=1.152 w
p3=0.768 w
pt=4.224 w
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
R = 1K||2K||3K = 6/11 K
Power P = V2/R = (48*18*11)/6K = 4.224 mA
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