Electronics - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 3)
3.
If the load in the given circuit is 120 k
, what is the loaded output voltage?
Discussion:
19 comments Page 1 of 2.
Amit said:
1 decade ago
RL is given as 120k
So 1st calculate the total resistance by
120k||8K=7500
then 2k+7500=9500
N
v=ir
20=i*9500
i=20/9500=2.1052exp-3
Now by current division rule
current through RL will be say
I2=i*8k/(8k+120k)=1.315exp-4
again
v=ir
So say voltage across RL be v2
v2=1.315exp-4*120k
=15.79v
same can be solved using voltage division rule
So 1st calculate the total resistance by
120k||8K=7500
then 2k+7500=9500
N
v=ir
20=i*9500
i=20/9500=2.1052exp-3
Now by current division rule
current through RL will be say
I2=i*8k/(8k+120k)=1.315exp-4
again
v=ir
So say voltage across RL be v2
v2=1.315exp-4*120k
=15.79v
same can be solved using voltage division rule
Rekha said:
1 decade ago
Very simple procedure,
Step 1:Find parallel combination of 8K & 120K = 7.5K
Step 2:Now 2k &7.5k are in series so total resistance=9.5k
Step 3:Find ct I=20v/9.5k= 2.105mA,
Step 4:find drop across 2K is 2.105m*2K=4.2V
Step 5:Finally Vl=20-4.2=15.8v.
Step 1:Find parallel combination of 8K & 120K = 7.5K
Step 2:Now 2k &7.5k are in series so total resistance=9.5k
Step 3:Find ct I=20v/9.5k= 2.105mA,
Step 4:find drop across 2K is 2.105m*2K=4.2V
Step 5:Finally Vl=20-4.2=15.8v.
Karthik said:
1 decade ago
In the Given Circuit RL=120K
So RL and R2 in parallel,
R2=8K
1/R=1/R2+1/RL
1/R=1/8+1/120
1/R=0.13333
R=7.5K
THEN R AND R1 IN SERIES SO TOTAL R =9.5K
CURRENT I =0.0021A
SO
VL = I*R VL=0.0021*7.5K
VL =15.75..
Thanks,
Karthik
So RL and R2 in parallel,
R2=8K
1/R=1/R2+1/RL
1/R=1/8+1/120
1/R=0.13333
R=7.5K
THEN R AND R1 IN SERIES SO TOTAL R =9.5K
CURRENT I =0.0021A
SO
VL = I*R VL=0.0021*7.5K
VL =15.75..
Thanks,
Karthik
(1)
Thiru said:
8 years ago
Can we choose this method?
Apply *voltage Divider Rule *
V=((Total voltage *opposite Resistance) /Total Resistance)
20*8/8+2=16V.
Whatever the Voltage across parallel is same as the other side That's why its 16v.
Apply *voltage Divider Rule *
V=((Total voltage *opposite Resistance) /Total Resistance)
20*8/8+2=16V.
Whatever the Voltage across parallel is same as the other side That's why its 16v.
Aravind said:
8 years ago
8k parallel 10k= (8*120)/(8+120)= 7.5k.
Now,
Voltage across 7.5k = Vol.across 120k (bz 8k parallel 120k)
So vol across 7.5k =(20V * 7.5k)/2k + 7.5k,
= 15.79V.
(Vol.divider equation)
Now,
Voltage across 7.5k = Vol.across 120k (bz 8k parallel 120k)
So vol across 7.5k =(20V * 7.5k)/2k + 7.5k,
= 15.79V.
(Vol.divider equation)
(5)
AKHIL VA said:
6 years ago
resistor 120k is parallel with 8k.
So r=1/120+1/8 = 7.5k.
vout=r2/(r1+r2) = 7.5/((2+7.5).
= 15.78947368421053 V,
= 15.79 V.
So r=1/120+1/8 = 7.5k.
vout=r2/(r1+r2) = 7.5/((2+7.5).
= 15.78947368421053 V,
= 15.79 V.
(5)
Ishwar said:
1 decade ago
Voltage division principle is used.
This way
First determine the parallel combination of R2 and RL and then find the voltage division between the resultant and R1.
This way
First determine the parallel combination of R2 and RL and then find the voltage division between the resultant and R1.
Vc_davv said:
8 years ago
Don't be so complex.
Use thevenin theorem.
So vth = 16 v and Rth = 1.6 kohm.
Voltage across Rl (load 120kohm) by VDL = (16*120K/120k+1.6k)=15.7v.
Use thevenin theorem.
So vth = 16 v and Rth = 1.6 kohm.
Voltage across Rl (load 120kohm) by VDL = (16*120K/120k+1.6k)=15.7v.
(1)
Madhusudan said:
1 decade ago
RL||R2=7.5k
and
vltg drop across 7.5k s
(7.5/(7.5+2))*20=15.79...
and
vltg drop across 7.5k s
(7.5/(7.5+2))*20=15.79...
Rajesh madavi said:
1 decade ago
8*120/8+120 =7.5k
v=20*7.5k/7.5k+2k
=20*15/19
=300/19
=15.79v
v=20*7.5k/7.5k+2k
=20*15/19
=300/19
=15.79v
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