Electronics - Series Circuits - Discussion

Discussion Forum : Series Circuits - General Questions (Q.No. 10)
10.

Calculate the voltage at point C in the given circuit.

+1.48 V
+0.94 V
+11.6 V
+3 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
12 comments Page 1 of 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.
(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.
(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.
(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.

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.

Kadarikota teja said:   9 years ago
It is not 4, it is 0.4.

Arunkumar said:   9 years ago
12v - 4v = 11.6 what is this?

Armandwish said:   10 years ago
Why 12-9 instead of 12+9? Please explain briefly and clearly.

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
(1)

Karthikkr said:   1 decade ago
V1 = 9*20/150 = 1.2

V2 = 12*130/150 = 10.4

VC = 10.4+1.2=11.6
(1)


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