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.
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)
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.
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.
Jamil hassan said:
8 years ago
Simply, we calculate the current.
By adding all the resistance and further calculate v : 12-9::: 3V...Then i is .02mA.
Voltage drop is .02* 20 = .4.
12 - .4 = 11.6.
By adding all the resistance and further calculate v : 12-9::: 3V...Then i is .02mA.
Voltage drop is .02* 20 = .4.
12 - .4 = 11.6.
(2)
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)
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
Sahal Waf said:
7 years ago
I=(12-9)/150 = 0.02mA,
The voltage at B = 9V + 0.02mA * (47+27+56)K ohm = 9V+2.6V = 11.6V.
The voltage at B = 9V + 0.02mA * (47+27+56)K ohm = 9V+2.6V = 11.6V.
(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
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)
Armandwish said:
10 years ago
Why 12-9 instead of 12+9? Please explain briefly and clearly.
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