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.
Pradeep said:
8 years ago
Only 5.7 is correct. I agree.
Santosh said:
9 years ago
The voltage will divide in series where the current is in parallel. In question indicated a voltage divider rule so resistors also in series.
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.
Archana said:
1 decade ago
But it has not said that o/p resistance parallel with one of 100k resistor.
(1)
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.
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.
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