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 2 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
(1)
Mahanthesh aadi said:
1 decade ago
Ok then answer is 10.48v, if we want to measure voltage at point A, and point C, then which method have to use? Can you explain me?
(1)
Adco said:
1 decade ago
Let B = voltage at node B.
(9-B)/(47k+27k) = (B-12)/(56k+20k).
B = 10.48A.
Equate them in their current because they are in series.
(9-B)/(47k+27k) = (B-12)/(56k+20k).
B = 10.48A.
Equate them in their current because they are in series.
Samuel44 said:
1 decade ago
From KVL,
I(R1+R2+R3+R4)+12-9 = 0.
150I = 3
I = 0.02A.
THEREFORE,
Vb = 12-I(R3+R4) Vb=voltage drop at B.
Vb = 12-0.02(56+20).
Vb = 0.02(76).
Vb = 10.48V.
I(R1+R2+R3+R4)+12-9 = 0.
150I = 3
I = 0.02A.
THEREFORE,
Vb = 12-I(R3+R4) Vb=voltage drop at B.
Vb = 12-0.02(56+20).
Vb = 0.02(76).
Vb = 10.48V.
SATYA GOUTHAM said:
1 decade ago
In the circuit the voltages are connected in reverse
Net voltage:
v=12-9 = 3;
Net resistance R = 27+47+20+56 = 150.
CURRENT I = 3/150=0.02AMPS.
Voltage at B = (47+27)0.02+9.
= 1.48+9..
= 10.48VOLTS
Net voltage:
v=12-9 = 3;
Net resistance R = 27+47+20+56 = 150.
CURRENT I = 3/150=0.02AMPS.
Voltage at B = (47+27)0.02+9.
= 1.48+9..
= 10.48VOLTS
Ken said:
1 decade ago
This is what I worked out
Where did I go wrong?
Rt = 1000(47+27+57+20) = 150000
Vs1 at a = (47/150)9V = 2.82V
Vs1 at b = ((47+27)/150)9V = 4.44V
Vs1 at c = ((47+27+56)/150)9V = 7.8V
Vs1 at d = (150/150)9V = 9V
Vs2 at c = (20/150)12V = 1.6V
Vs2 at b = ((20+56)/150)12V = 6.08V
Vs2 at a = ((20+56+27)/150)12V = 8.24
Vs2 - Vs1 = 3V
Vs1 - Vs2 = -3V
Vs1 + Vs2 = 21V
Vs1 + Vs2 at b = 10.52?
Where did I go wrong?
Rt = 1000(47+27+57+20) = 150000
Vs1 at a = (47/150)9V = 2.82V
Vs1 at b = ((47+27)/150)9V = 4.44V
Vs1 at c = ((47+27+56)/150)9V = 7.8V
Vs1 at d = (150/150)9V = 9V
Vs2 at c = (20/150)12V = 1.6V
Vs2 at b = ((20+56)/150)12V = 6.08V
Vs2 at a = ((20+56+27)/150)12V = 8.24
Vs2 - Vs1 = 3V
Vs1 - Vs2 = -3V
Vs1 + Vs2 = 21V
Vs1 + Vs2 at b = 10.52?
Ishpreet singh said:
10 years ago
If we have to find the voltage at point c then voltages 9v and 12v are added/or subtract?
Amy Jack said:
10 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.
Jason druye said:
9 years ago
What's superposition theorem?
Poor mar said:
2 years ago
Use Millman's Theorem:
This is used to find the voltage across parallel branches with different voltage sources.
Vb = (9+73K)+(12+76K)/(1/73K)+(1/76K).
This is used to find the voltage across parallel branches with different voltage sources.
Vb = (9+73K)+(12+76K)/(1/73K)+(1/76K).
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