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 3 of 3.
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
Anand said:
1 decade ago
@Rohit can you explain about the answer.
Rohit said:
1 decade ago
The above method is 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
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