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.
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.
Shivi said:
1 decade ago
I=(12-9)/150=.02mA
starting frm 1 direction
Now,
12-Vc=(.02*10^-3)(20*10^3)
So, Vc=11.6
Similarly,
Vc-Vb=(.02*10^-3)(56*10^3)
11.6-Vb=1.12
So, Vb=10.48V
starting frm 1 direction
Now,
12-Vc=(.02*10^-3)(20*10^3)
So, Vc=11.6
Similarly,
Vc-Vb=(.02*10^-3)(56*10^3)
11.6-Vb=1.12
So, Vb=10.48V
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).
Ross said:
2 years ago
The answer should be approximately 1.5 volts the power sources positive facing positive cancel out resulting in a source voltage of 3 volts.
(4)
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.
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)
Rishi pandey said:
1 decade ago
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
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
Bhanu said:
1 decade ago
Here the net voltage should be 12-9 or 9-12. Why.
Because both the facing polarities are same.
Because both the facing polarities are same.
Babu said:
1 decade ago
Applying KVL,
9+150I-10=0
I=0.02 A.
V at B=12-I*(20+56)
=10.48 v.
I flow from C to A
9+150I-10=0
I=0.02 A.
V at B=12-I*(20+56)
=10.48 v.
I flow from C to A
Atif said:
1 decade ago
The batteries are connected in series opposing configuration, Then how voltages can b add?
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