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 2 of 3.
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?
Pradeep said:
8 years ago
Only 5.7 is correct. I agree.
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
Pranjal singh said:
9 years ago
But it's not given in the question that resistance is series or parallel. Then how to solve this.
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.
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.
Shoji said:
1 decade ago
Explanation by @Cpreddy seems to be correct approach to the problem.
Maya said:
1 decade ago
One 100k and 1M are in parallel and other one is in series then apply VDR.
Uduak Ekanem said:
1 decade ago
But it is not stated in the question if the resistors are connected in series or parallel.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers