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.
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)
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)
Faizan said:
7 years ago
v = i * r = 28/7 * 4.8 = 19.2.
(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)
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)
Manohar said:
1 decade ago
V = I*R.
Given V = 28V.
R equivalent = 7K.
Therefore I = V/R = 4mA.
Res from B to D = 4.8K.
V from B to D = I*Res =4*4.8*10^3*10^-3 = 19.2.
Given V = 28V.
R equivalent = 7K.
Therefore I = V/R = 4mA.
Res from B to D = 4.8K.
V from B to D = I*Res =4*4.8*10^3*10^-3 = 19.2.
Gopal Rana said:
1 decade ago
V=IR then total resistance R=(R1+R2+R3).
Then R=(2.2+3.3+1.5), R=7Kohm, V=IR, I=28/7, I=4ma.
Now points b and d the total resistane Rbd=3.3+1.5=4.8kohm.
Now V=IR,4*4.8=19.2.
Am I Right?
Then R=(2.2+3.3+1.5), R=7Kohm, V=IR, I=28/7, I=4ma.
Now points b and d the total resistane Rbd=3.3+1.5=4.8kohm.
Now V=IR,4*4.8=19.2.
Am I Right?
Ahammed Najeeb said:
1 decade ago
Dear mats,
we have,
R1=2.2kohms,R2=3.3Kohms,R3=1.5Kohms,
convert it into ohms,then,
R1=2200ohms,R2=3300Ohms,R3=1500Ohms,R=(R1+R2+R3)=7000 Ohms
V=28v,
so I from the wquation V=IR,
I=V/R,
I=28v/7000 ohms,
I=0.004A,
And we have to find the voltage betwean B and D,
Resisters b/w B&D are R2&R3,So
V b/w B&D is V=I(R2+R3),
V=0.004(3300ohms+1500ohms),
V=19.2v.
we have,
R1=2.2kohms,R2=3.3Kohms,R3=1.5Kohms,
convert it into ohms,then,
R1=2200ohms,R2=3300Ohms,R3=1500Ohms,R=(R1+R2+R3)=7000 Ohms
V=28v,
so I from the wquation V=IR,
I=V/R,
I=28v/7000 ohms,
I=0.004A,
And we have to find the voltage betwean B and D,
Resisters b/w B&D are R2&R3,So
V b/w B&D is V=I(R2+R3),
V=0.004(3300ohms+1500ohms),
V=19.2v.
Arun Jain 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
OR
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
V= (R1 + R2) (Total Voltage/(R1+R2+R3))
V= (3.3K + 1.5K) (28/(2.2K + 3.2K + 1.5K))
V=19.2 V
OR
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
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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