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.
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)
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)
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)
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)
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
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.
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