Electronics - Series Circuits - Discussion

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

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

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=V/R
=21/150
=0.14
V=I * R
= 0.14 * 47+27
= 10.36
is approximately near to 10.48

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.

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.

According to KVL.

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

vb=10.48.

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.

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?