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Series-Parallel Circuits

Q1: If the load in the given circuit is 80 komega.gif , what is the bleeder current?

A 196 mu.gif A

B 1.96 mA

C 2 mA

D 2.16 mA

ANS:B - 1.96 mA

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).



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