Electronics - Series Circuits - Discussion
Discussion Forum : Series Circuits - General Questions (Q.No. 7)
7.
Calculate the voltage at point B in the given circuit.
Discussion:
36 comments Page 3 of 4.
Vaishali said:
1 decade ago
I=V/R
=21/150
=0.14
V=I * R
= 0.14 * 47+27
= 10.36
is approximately near to 10.48
=21/150
=0.14
V=I * R
= 0.14 * 47+27
= 10.36
is approximately near to 10.48
Ishpreet singh said:
9 years ago
If we have to find the voltage at point c then voltages 9v and 12v are added/or subtract?
Quiet Kid said:
2 years ago
Vr 9V = (76k/150k) .9 = 4.56.
Vr 12V = (74k/150k) .12 = 5.92.
So, 4.56 + 5.92 = 10.48.
Vr 12V = (74k/150k) .12 = 5.92.
So, 4.56 + 5.92 = 10.48.
(7)
Amy Jack said:
9 years ago
By using Superposition theorem we can solve this.
I got the correct answer as 10.5.
I got the correct answer as 10.5.
Anonymous said:
9 years ago
KVL is best & short method.
((vb-9)/74) = ((12-vb)/76).
vb = 10.48.
((vb-9)/74) = ((12-vb)/76).
vb = 10.48.
(6)
Rizwan said:
1 decade ago
Please Explain how the voltage of these batteries adding?
Aneeshrajan said:
1 decade ago
I=21/150
V1=74*21/150
V2=76*21/150
V at B=(V1+V2)/2=10.5
V1=74*21/150
V2=76*21/150
V at B=(V1+V2)/2=10.5
Faizan said:
7 years ago
i1 = 9/150,
i2 = 12/150,
vb = i1*74 + i2*76,
vb = 10.52.
i2 = 12/150,
vb = i1*74 + i2*76,
vb = 10.52.
(1)
Aaadi said:
1 decade ago
Well!
The best thing is to apply Law of superposition.
The best thing is to apply Law of superposition.
(1)
Piyush said:
1 decade ago
According to KVL.
((vb-9)/74)=((12-vb)/76).
vb=10.48.
((vb-9)/74)=((12-vb)/76).
vb=10.48.
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