Electronics - Series Circuits - Discussion
Discussion Forum : Series Circuits - General Questions (Q.No. 10)
10.
Calculate the voltage at point C in the given circuit.
Discussion:
12 comments Page 1 of 2.
Rishi pandey said:
1 decade ago
I=(12-9)/150=0.02mA
voltage at B =9V + 0.02mA*(47+27+56)K ohm= 9V+2.6V= 11.6V
voltage at B =9V + 0.02mA*(47+27+56)K ohm= 9V+2.6V= 11.6V
Arun Jain said:
1 decade ago
I=(12-9)/150=0.02mA
voltage at C =9V + 0.02mA*(47+27+56)K ohm= 9V+2.6V= 11.6V
OR
voltage at C =12V - 0.02mA*(20)K ohm= 12V-.4V= 11.6V
voltage at C =9V + 0.02mA*(47+27+56)K ohm= 9V+2.6V= 11.6V
OR
voltage at C =12V - 0.02mA*(20)K ohm= 12V-.4V= 11.6V
Karthikkr said:
1 decade ago
V1 = 9*20/150 = 1.2
V2 = 12*130/150 = 10.4
VC = 10.4+1.2=11.6
V2 = 12*130/150 = 10.4
VC = 10.4+1.2=11.6
(1)
Suresh Sharma said:
1 decade ago
I=(12-9)/150=0.02mA
voltage at C =9V + 0.02mA*(47+27+56)k ohm = 9V+2.6V= 11.6V
OR
voltage at C =12V - 0.02mA*(20)K ohm= 12V-.4V= 11.6V is the right answer
voltage at C =9V + 0.02mA*(47+27+56)k ohm = 9V+2.6V= 11.6V
OR
voltage at C =12V - 0.02mA*(20)K ohm= 12V-.4V= 11.6V is the right answer
(1)
Armandwish said:
10 years ago
Why 12-9 instead of 12+9? Please explain briefly and clearly.
Arunkumar said:
9 years ago
12v - 4v = 11.6 what is this?
Kadarikota teja said:
9 years ago
It is not 4, it is 0.4.
Neeraj said:
9 years ago
@Armandwish.
12-9 because both the voltage source has the same polarity and hence their net potential is the difference between them. For a further explanation, you can see it by applying KVL to the circuit.
12-9 because both the voltage source has the same polarity and hence their net potential is the difference between them. For a further explanation, you can see it by applying KVL to the circuit.
Neeraj said:
9 years ago
@Armandwish,
12-9 because both the voltage source has the same polarity and hence their net potential is the difference between them. For the further explanation, you can see it by applying KVL to the circuit.
12-9 because both the voltage source has the same polarity and hence their net potential is the difference between them. For the further explanation, you can see it by applying KVL to the circuit.
KIRAN V said:
9 years ago
Apply KVL,
(Vc - 9V)/(47 K Ohm + 27 K Ohm + 56 K Ohm) = (Vc - 12 V)/(20 K Ohm).
(Vc - 9V)/(130 K Ohm) = (Vc - 12 V)/(20 K Ohm).
By Cross-multiplying we get,
(130 K Ohm) * (Vc - 12 V) = (20 K Ohm) * (Vc - 9 V).
Finally we get,
Vc = (1740 * 10^3) / (150* 10^3).
Vc = 11.6 V.
(Vc - 9V)/(47 K Ohm + 27 K Ohm + 56 K Ohm) = (Vc - 12 V)/(20 K Ohm).
(Vc - 9V)/(130 K Ohm) = (Vc - 12 V)/(20 K Ohm).
By Cross-multiplying we get,
(130 K Ohm) * (Vc - 12 V) = (20 K Ohm) * (Vc - 9 V).
Finally we get,
Vc = (1740 * 10^3) / (150* 10^3).
Vc = 11.6 V.
(1)
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