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.
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
Rohit said:
1 decade ago
The above method is wrong.
Anand said:
1 decade ago
@Rohit can you explain about the answer.
CPREDDY said:
1 decade ago
R1=100K,R2=100K,R3=1000K
HERE R1,R2 ARE IN SERIES WITH 12V.
AND R2,R3 ARE PARALLEL
VOUT=12*(100||1000)/(100+100||1000)=120/21=5.7
HERE R1,R2 ARE IN SERIES WITH 12V.
AND R2,R3 ARE PARALLEL
VOUT=12*(100||1000)/(100+100||1000)=120/21=5.7
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.
Uduak Ekanem said:
1 decade ago
But it is not stated in the question if the resistors are connected in series or parallel.
Maya said:
1 decade ago
One 100k and 1M are in parallel and other one is in series then apply VDR.
Shoji said:
1 decade ago
Explanation by @Cpreddy seems to be correct approach to the problem.
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.
Archana said:
1 decade ago
But it has not said that o/p resistance parallel with one of 100k resistor.
(1)
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