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.
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?
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)
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?
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
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.
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.
Ganesha said:
1 decade ago
1. Here the net voltage should be 12-9 or 9-12. Why.
Because both the facing polarities are same.
2. The batteries are connected in series opposing configuration, Then how voltages can be add?
3. Please Explain how the voltage of these batteries adding?
Because both the facing polarities are same.
2. The batteries are connected in series opposing configuration, Then how voltages can be add?
3. Please Explain how the voltage of these batteries adding?
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.
Piyush said:
1 decade ago
According to KVL.
((vb-9)/74)=((12-vb)/76).
vb=10.48.
((vb-9)/74)=((12-vb)/76).
vb=10.48.
Vishal Jariwala said:
1 decade ago
First of all find out value of current.
I=0.02ma.
Find all voltages across resistor using KVL.
V47=0.94v, V27=0.54v, V56=1.12v, V20=0.4v.
So,
Voltage at B w.r.t ground=12-(v47+v27)=12-(1.12+.40)=10.48volts.
I=0.02ma.
Find all voltages across resistor using KVL.
V47=0.94v, V27=0.54v, V56=1.12v, V20=0.4v.
So,
Voltage at B w.r.t ground=12-(v47+v27)=12-(1.12+.40)=10.48volts.
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