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 3 of 4.
Ganesha said:
1 decade ago
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?
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?
Aneeshrajan said:
1 decade ago
I=21/150
V1=74*21/150
V2=76*21/150
V at B=(V1+V2)/2=10.5
V1=74*21/150
V2=76*21/150
V at B=(V1+V2)/2=10.5
Piyush said:
1 decade ago
According to KVL.
((vb-9)/74)=((12-vb)/76).
vb=10.48.
((vb-9)/74)=((12-vb)/76).
vb=10.48.
Vishal Jariwala said:
1 decade ago
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.
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.
Kiran Hatti said:
1 decade ago
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.
Vaishali said:
1 decade ago
I=V/R
=21/150
=0.14
V=I * R
= 0.14 * 47+27
= 10.36
is approximately near to 10.48
=21/150
=0.14
V=I * R
= 0.14 * 47+27
= 10.36
is approximately near to 10.48
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
Shekar said:
1 decade ago
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.
So voltage at B is greater than 9.
In options we have A only >9.
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
Wajid said:
1 decade ago
Guys use superposition theorem to solve the problem.
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