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 3 of 4.
Gautam said:
1 decade ago
In this circuit both the voltage sources are connected in reverse mode to each other so,
I=(12-9)/150= 0.02mA.
Let 9V as GND then Potential at B= (47+27)K * 0.02mA =1.48V,
So pot. at B if GND is 0V =9V+1.48V =10.48V Answer.
Else let 12V as GND point then B = (56+20)K * 0.02mA = 1.52V.
So Pot. at B is = 12V - 1.52V = 10.48V Answer.
I=(12-9)/150= 0.02mA.
Let 9V as GND then Potential at B= (47+27)K * 0.02mA =1.48V,
So pot. at B if GND is 0V =9V+1.48V =10.48V Answer.
Else let 12V as GND point then B = (56+20)K * 0.02mA = 1.52V.
So Pot. at B is = 12V - 1.52V = 10.48V Answer.
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.
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?
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)
Ishpreet singh said:
9 years ago
If we have to find the voltage at point c then voltages 9v and 12v are added/or subtract?
Amy Jack said:
9 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?
Anonymous said:
9 years ago
KVL is best & short method.
((vb-9)/74) = ((12-vb)/76).
vb = 10.48.
((vb-9)/74) = ((12-vb)/76).
vb = 10.48.
(6)
Jonathan said:
8 years ago
Using Superposition:
Total Resistance = 150kOhm
@12V = On, 9V = Off:
I=80uA
Voltage Drop at C = 10.4V
Voltage Drop at B = 5.92V (save for later)
@12V = Off, 9V = On:
I =60uA
Voltage drop at A = 6.18V
Voltage drop at B = 4.56V (save for later)
solving for total drop at B
Vbtotal = Vb + Vb` = 5.92V + 4.56V = 10.48V.
Total Resistance = 150kOhm
@12V = On, 9V = Off:
I=80uA
Voltage Drop at C = 10.4V
Voltage Drop at B = 5.92V (save for later)
@12V = Off, 9V = On:
I =60uA
Voltage drop at A = 6.18V
Voltage drop at B = 4.56V (save for later)
solving for total drop at B
Vbtotal = Vb + Vb` = 5.92V + 4.56V = 10.48V.
(7)
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