Electronics - Series Circuits - Discussion

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

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

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?

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.

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.

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?

Different voltages in series will add each other. The total resistance at Right side of point 'B' is 76k and the total resistance at left side of Point 'B' is almost 76k. So voltage at point 'B' is exactly average of both voltages.

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

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.

By seeing the current direction. Voltage at point B is greater than voltage at point A and ground after A voltage is 9v.

So voltage at B is greater than 9.

In options we have A only >9.