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.
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)
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