Electronics - Series Circuits - Discussion

Vr 9V = (76k/150k) .9 = 4.56.
Vr 12V = (74k/150k) .12 = 5.92.
So, 4.56 + 5.92 = 10.48.

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.

KVL is best & short method.

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

vb = 10.48.

The answer should be approximately 1.5 volts the power sources positive facing positive cancel out resulting in a source voltage of 3 volts.

This is a situation where Two water tanks are connected each other with a single pipe one has More water and Other has less water.

So water(Current) flow from higher to lower.

Here V2=12V and V1=9V (V2>V1) so current flow from V2 to V1

I=Veff/Rtot

Veff=V2-V1 (V1&V2 opposite to each other & V2>V1)

Veff=12-9=3V

Rtot= R1+R2+R3+R4=150kOhms

I=3/150 x10^-3

I=0.02x10^-3 A

VB = V2 - (VR3+VR4)

VB = V2 - .02x10^-3(R3+R4) (Since current is equal in series circuit)

VB = 12 - 0.02x10^-3x76x10^3

VB = 12-1.52

VB = 10.48 V

Why R1 and R2 are not considered for voltgae drop because no current from V1 to V2

I cannot understand this, please help me to get this.

i1 = 9/150,
i2 = 12/150,
vb = i1*74 + i2*76,
vb = 10.52.

Can anyone explain detail about superposition theorem?

Well!

The best thing is to apply Law of superposition.

Since batteries are connected opposing each other the current in the circuit will be from higher voltage to lower.

It is just like 2 water tanks connected to each other with a single pipe.

Total current in circuit I=Effective voltage/Total Resistance

Effective Voltage = V2-V1 (since V2>V1&connected opposite to each other)
Total Resistance = R1+R2+R3+R4 (since all are in series)

I=(3/150)x10^-3
I=0.02x10^-3 A

So current flow is from V2 to V1

VB = V2-(VR3+VR4) (since current is equal in series circuit)
VB = 12 - (0.02x10^-3x20x10^3+0.02x10^-3x56x10^3)
VB= 12 - 0.02X76
VB = 12 - 1.52
VB = 10.48 V

Why VB is Measure only with respect to V2 because there is no current from V1 towards VB so VB is only drop across R3&R4