Electronics - Series Circuits - Discussion
Discussion Forum : Series Circuits - General Questions (Q.No. 9)
9.
What is the voltage at points B to D in the given circuit?
Discussion:
14 comments Page 2 of 2.
Saichaitanya said:
1 decade ago
As per the voltage Divider rule:
V= (R1 + R2) (Total Voltage/(R1+R2+R3))
V= (3.3K + 1.5K) (28/(2.2K + 3.2K + 1.5K))
V=19.2 V
V= (R1 + R2) (Total Voltage/(R1+R2+R3))
V= (3.3K + 1.5K) (28/(2.2K + 3.2K + 1.5K))
V=19.2 V
Dhaval Makhecha said:
1 decade ago
V=28v
R=R1+R2+R3=7kohm
I=V/R= 28/7= 4 ma.
Now, voltage between point B and D,
V=I(R2+R3) = 4(3.3+1.5) = 4*4.8 = 19.2v
So, V=19.2v
R=R1+R2+R3=7kohm
I=V/R= 28/7= 4 ma.
Now, voltage between point B and D,
V=I(R2+R3) = 4(3.3+1.5) = 4*4.8 = 19.2v
So, V=19.2v
Vinod kumar yadav said:
1 decade ago
@Anil Kamani:
Sorry, you are wrong becouse current will flow from B-C-D path not from B-A-D. If u want to flow the the current from B to D through point A then substract the 8.8v from 28V due opposite direction of current.
So, u will get
V=-(8.8-28)v
V=+19.2 v
Sorry, you are wrong becouse current will flow from B-C-D path not from B-A-D. If u want to flow the the current from B to D through point A then substract the 8.8v from 28V due opposite direction of current.
So, u will get
V=-(8.8-28)v
V=+19.2 v
Anil kamani said:
1 decade ago
Hi,
I think the calculation shall take as below.
Addition of all resister = 7 kohms.
Voltage source = 28 volt. So current flow = 4 ma.
==> voltage drop accross r1 = 4*2.2 = 8.8 volt.
Am I correct. ?
I think the calculation shall take as below.
Addition of all resister = 7 kohms.
Voltage source = 28 volt. So current flow = 4 ma.
==> voltage drop accross r1 = 4*2.2 = 8.8 volt.
Am I correct. ?
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