Electronics - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 10)
10.
If the load in the given circuit is 80 k
, what is the bleeder current?
196 
A
A1.96 mA
2 mA
2.16 mA
Discussion:
7 comments Page 1 of 1.
Ash said:
4 years ago
As R2 AND Rl are parallel.
> 80x8/80 + 8 =7.2 kohm.
As R1 is in series with the above combination,
>7.2 kohm + 2kohm =9.2 kohm.
Total current I=V/R =20/9. 2 Kohm =2. 173 ma.
Next, current through R2= Ix Rl / Rl +R2 = 1.976 = 1.96 (approximated to).
> 80x8/80 + 8 =7.2 kohm.
As R1 is in series with the above combination,
>7.2 kohm + 2kohm =9.2 kohm.
Total current I=V/R =20/9. 2 Kohm =2. 173 ma.
Next, current through R2= Ix Rl / Rl +R2 = 1.976 = 1.96 (approximated to).
Manoj said:
7 years ago
Here, RL + R2 = 88.
Paresh said:
8 years ago
Why we are use 88?
How to come please anyone give me explanation?
How to come please anyone give me explanation?
Ameer said:
9 years ago
Nice @Sanj.
Faisal amin said:
1 decade ago
Here bleeder current is the current on R2 in the absence of load resistor RL. It means bleeder current is the unloaded current on the specific resistor where load is to be connected.
(1)
V.Nagendran said:
1 decade ago
Can you please the Bleeder current?
Sanj said:
1 decade ago
Find the total resistance which will be = 9.2kohm
Total current is =2.15mV
Bleeder current is current through resistor R2 which is= 1.96mA
Use the formula current through r2= rL/(r2+rL)
(80/88)*2.15 = 1.95454545
Total current is =2.15mV
Bleeder current is current through resistor R2 which is= 1.96mA
Use the formula current through r2= rL/(r2+rL)
(80/88)*2.15 = 1.95454545
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers