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 1 of 4.
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
Bhanu said:
1 decade ago
Here the net voltage should be 12-9 or 9-12. Why.
Because both the facing polarities are same.
Because both the facing polarities are same.
Atif said:
1 decade ago
The batteries are connected in series opposing configuration, Then how voltages can b add?
Rizwan said:
1 decade ago
Please Explain how the voltage of these batteries adding?
Rishi pandey said:
1 decade ago
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
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
Gopiram Verma said:
1 decade ago
Ya Rishi is right.
Faiz Ullah said:
1 decade ago
Here voltages are not added its subtracted 12V-9V=3V
Wajid said:
1 decade ago
Guys use superposition theorem to solve the problem.
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
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.
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