Electronics - Series Circuits - Discussion

Discussion Forum : Series Circuits - General Questions (Q.No. 7)
7.

Calculate the voltage at point B in the given circuit.

+10.48 V
+0.94 V
+2.6 V
+3 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
36 comments Page 1 of 4.

Wil said:   1 year ago
I cannot understand this, please help me to get this.
(1)

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).

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)

Quiet Kid said:   2 years ago
Vr 9V = (76k/150k) .9 = 4.56.
Vr 12V = (74k/150k) .12 = 5.92.
So, 4.56 + 5.92 = 10.48.
(7)

Faizan said:   7 years ago
i1 = 9/150,
i2 = 12/150,
vb = i1*74 + i2*76,
vb = 10.52.
(1)

Balu said:   7 years ago
Can anyone explain detail about superposition theorem?
(1)

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.
(7)

Anonymous said:   9 years ago
KVL is best & short method.

((vb-9)/74) = ((12-vb)/76).

vb = 10.48.
(6)

Jason druye said:   9 years ago
What's superposition theorem?

Amy Jack said:   9 years ago
By using Superposition theorem we can solve this.

I got the correct answer as 10.5.


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