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 1 of 2.
Lucky said:
10 years ago
v = i*r.
So i = v/r.
Now here r1, r2, r3 are in series, so r = 7 oh.
i = 28/7.
So that vtg drop at a to b point is.
v = i*r.
= 4*2.2 = 8.8 v.
And vtg drop at point b to d is.
3.3 + 1.5 = 4.8 ohm.
v = i*r.
= 4*4.8.
= 19.2 v.
So i = v/r.
Now here r1, r2, r3 are in series, so r = 7 oh.
i = 28/7.
So that vtg drop at a to b point is.
v = i*r.
= 4*2.2 = 8.8 v.
And vtg drop at point b to d is.
3.3 + 1.5 = 4.8 ohm.
v = i*r.
= 4*4.8.
= 19.2 v.
(3)
Faizan said:
7 years ago
v = i * r = 28/7 * 4.8 = 19.2.
(2)
Karthikeyan said:
6 years ago
Hi guys,
There is another way to find the correct answer,
That is voltage divider rule V(find) = V(total) * (R2+R3)/(R1+R2+R3)
V(find) = 28 * (4.8k)/(7k) = 19.2v.
There is another way to find the correct answer,
That is voltage divider rule V(find) = V(total) * (R2+R3)/(R1+R2+R3)
V(find) = 28 * (4.8k)/(7k) = 19.2v.
(2)
KIRAN V said:
5 years ago
V = 28v.
R = R1+R2+R3 = 7k Ohm.
Va = Vin (R1/Req).
Va = 28 (2.2 K / 7 K),
Va = 8.8 V.
Voltage at points B to D.
Vtotal = Va + (Vb+Vc+Vd).
(Vb+Vc+Vd) = Vtotal - Va,
(Vb+Vc+Vd) = 28 - 8.8,
(Vb+Vc+Vd) = + 19.2 V.
R = R1+R2+R3 = 7k Ohm.
Va = Vin (R1/Req).
Va = 28 (2.2 K / 7 K),
Va = 8.8 V.
Voltage at points B to D.
Vtotal = Va + (Vb+Vc+Vd).
(Vb+Vc+Vd) = Vtotal - Va,
(Vb+Vc+Vd) = 28 - 8.8,
(Vb+Vc+Vd) = + 19.2 V.
(2)
Vijayakumar said:
1 decade ago
V = 28v.
R = R1+R2+R3 = 7kohm.
I = V/R = 28/7 = 4 ma.
Vb = iR2 = 4*3.3 = 13.2.
Vd = iR3 = 4*1.5 = 6.
V = Vb+Vd = 13.2+6 = 19.2V.
R = R1+R2+R3 = 7kohm.
I = V/R = 28/7 = 4 ma.
Vb = iR2 = 4*3.3 = 13.2.
Vd = iR3 = 4*1.5 = 6.
V = Vb+Vd = 13.2+6 = 19.2V.
(1)
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. ?
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
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
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
Naseem Ul Aziz said:
1 decade ago
We should calculate first:
it(total current)=vt/rt
vt=28 v
rt=7000 ohm
it=28/7000=0.004
Resistance from b to d=4800 ohm
v(b to d)=i*r(b to d)
i in series = total current
v(b to d)=0.004*4800=19.2 volt
Naseem Ul Aziz From Pakistan
it(total current)=vt/rt
vt=28 v
rt=7000 ohm
it=28/7000=0.004
Resistance from b to d=4800 ohm
v(b to d)=i*r(b to d)
i in series = total current
v(b to d)=0.004*4800=19.2 volt
Naseem Ul Aziz From Pakistan
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