Electrical Engineering - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 21)
21.
A voltage divider consists of two 100 k
resistors and a 12 V source. What will the output voltage be if a load resistor of 1 M is connected to the output?
resistors and a 12 V source. What will the output voltage be if a load resistor of 1 M is connected to the output?Discussion:
24 comments Page 1 of 3.
Subodh said:
2 years ago
Can anybody explain it properly?
(1)
Geeta H said:
3 years ago
@Jeyaraj
The vdr formula is;
Vout = R2 * Vin/(R1+R2).
Then you will get 5.7 as the answer.
So, here D is the right answer.
The vdr formula is;
Vout = R2 * Vin/(R1+R2).
Then you will get 5.7 as the answer.
So, here D is the right answer.
(1)
Kvnblz said:
3 years ago
By approximation method.
Vb = 12(100/200) = 6 V.
Assuming Germanium transistor,
Vb = 0.3 + Vo, thus Vo = 6-0.3 = 5.7V.
Vb = 12(100/200) = 6 V.
Assuming Germanium transistor,
Vb = 0.3 + Vo, thus Vo = 6-0.3 = 5.7V.
(1)
Raja Tayyab said:
4 years ago
Let me explain clearly in simple words. Firstly the circuit was 2 100k resistors in series and 12V source. Then a load resistor of 1M was added and load is always added in parallel to the source. This 1M will be parallel to one 100k. So the circuit becomes:
R1+ (R2||R3)= 100k+ (100k||1000K),
The voltage will be divided in 100k and 100k||1000k. So applying VDR at load.
100k||1000k=90k,
Vo = 12*[(90k)/(90k+100k)] = 5.7V.
R1+ (R2||R3)= 100k+ (100k||1000K),
The voltage will be divided in 100k and 100k||1000k. So applying VDR at load.
100k||1000k=90k,
Vo = 12*[(90k)/(90k+100k)] = 5.7V.
(7)
Adel said:
5 years ago
R1=100k , R2=100k , R3=1000k.
According to question,
R2||R3 i-e 100k||1000k = 90k.
Now R1=100k and R2 become 90k, now apply VDR.
Vo = 5.7 v.
According to question,
R2||R3 i-e 100k||1000k = 90k.
Now R1=100k and R2 become 90k, now apply VDR.
Vo = 5.7 v.
Gaurav Kumar said:
5 years ago
When it says, there is a voltage divider circuit then it means R's are in series otherwise it wouldn't be a V Divider ckt. And if Load is in Parallel it doesn't make sense.
Try to draw a ckt with R1 and R2 in series and Load in parallel load will short ckt in this case.
Try to draw a ckt with R1 and R2 in series and Load in parallel load will short ckt in this case.
HAris said:
5 years ago
@Cpreddy. You are wrong at the end how can you put 12V in VDR with Kohm resistor, units must be same of all the values while putting in any formula, You have to convert either voltage in to KV or Kohm into Ohm.
Rajendra said:
6 years ago
First one 100k ohm is parallel with 1M ohm solve it. You will get 90.9k ohms
Then use voltage divider formula, vout = vin x R2/R1+R2.
You will get 5.64 v which is equal to 5.7 volt. So the answer is option D.
Then use voltage divider formula, vout = vin x R2/R1+R2.
You will get 5.64 v which is equal to 5.7 volt. So the answer is option D.
Ishmael said:
7 years ago
@Dayanand.
When they say 'a load resistor of 1 Mega Ohm is connected to the output', I think you're meant to assume that it is therefore added in parallel i.e. the 'Voltage out' is measured across the parallel combination of 100k and 1000k. Hope that makes sense.
Correct me if I'm wrong.
When they say 'a load resistor of 1 Mega Ohm is connected to the output', I think you're meant to assume that it is therefore added in parallel i.e. the 'Voltage out' is measured across the parallel combination of 100k and 1000k. Hope that makes sense.
Correct me if I'm wrong.
Dayanand said:
8 years ago
They have not mentioned that 1M ohm is in parallel.
Then, how did you solve?
Then, how did you solve?
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