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.
Raja Tayyab said:
5 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)
Hanumantha Hk said:
1 decade ago
Hi, guys.
According to question r1 and r2 series than load res parallel with r2. ok.
r1 = 100k; r2 = 100k; RL = 1000k.
R2//RL = r2*RL/r2+RL = 90.
9k that's series with r1. So 90.9k+100k = 190.9k.
Total current I = V/R = 12/190.9K = 0.0628ma.
According current division rule. I2 = I r1/r1+RL = 0.0628m*100k/100k+1000k = 0.0057ma.
So finally we need across load VL = i2*RL = 0.0057ma*1000K = 5.7V.
According to question r1 and r2 series than load res parallel with r2. ok.
r1 = 100k; r2 = 100k; RL = 1000k.
R2//RL = r2*RL/r2+RL = 90.
9k that's series with r1. So 90.9k+100k = 190.9k.
Total current I = V/R = 12/190.9K = 0.0628ma.
According current division rule. I2 = I r1/r1+RL = 0.0628m*100k/100k+1000k = 0.0057ma.
So finally we need across load VL = i2*RL = 0.0057ma*1000K = 5.7V.
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.
JEYARAJ N said:
1 decade ago
Ans: 6.285 V
Explanation:
First We solve,
Parallal 100K and 1M ohm
1/R=1/R1+1/R2
1/R =1/100K + 1/1M
R=1/(0.00001 + 0.000001)
R=90909.09 ohm
In voltage devider circuit.
R1= 100KOhm
R2=90909.09 ohm
V=12V
Vout= R1/(R1+R2) * Vout
Vout= (100 K /(100K + 90909.09)) * 12
Vout= 6.285 V
Explanation:
First We solve,
Parallal 100K and 1M ohm
1/R=1/R1+1/R2
1/R =1/100K + 1/1M
R=1/(0.00001 + 0.000001)
R=90909.09 ohm
In voltage devider circuit.
R1= 100KOhm
R2=90909.09 ohm
V=12V
Vout= R1/(R1+R2) * Vout
Vout= (100 K /(100K + 90909.09)) * 12
Vout= 6.285 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.
Henry said:
1 decade ago
Vt=12v, Rt=200k (from R1(100K)+R2(100K)), we can get It=0.00006amps.
Solving for Vout, Vout= It x R'.
R' = (100K)(1M)/(100K)+(1M) = 90909.09091 ohms.
so, Vout = (0.00006)x(90909.09091) = 5.45 volts.
Answer: Option D: 5.7volts.
Solving for Vout, Vout= It x R'.
R' = (100K)(1M)/(100K)+(1M) = 90909.09091 ohms.
so, Vout = (0.00006)x(90909.09091) = 5.45 volts.
Answer: Option D: 5.7volts.
HAris said:
6 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:
7 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.
OLINI said:
10 years ago
R1 = 100 k; R2 = 100 k; RL = 1000 k.
R2//RL = (100 k * 1000 k)/(100 k + 1000 k) = 90.909 k.
By voltage divider.
Vout1 = 12* ((100 k)/(100 k + 90.909 k)) = 6.2857 V.
Vout 2 = 12 - 6.2857 = 5.7 V.
R2//RL = (100 k * 1000 k)/(100 k + 1000 k) = 90.909 k.
By voltage divider.
Vout1 = 12* ((100 k)/(100 k + 90.909 k)) = 6.2857 V.
Vout 2 = 12 - 6.2857 = 5.7 V.
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.
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