prove that series combination resistance have maximum and parallel combination resistance have minimum current

Asked by  | 23rd Sep, 2012, 08:08: PM

Expert Answer:

When two res are connected in series:
eq res:R=r1+r2
current in the ckt:   Is=V/(r1+r2)........(1)
now R ie (r1+r2)sh always be greater than individual res r1 or r2.
so the current will obviously be lesser in comparison with the situation when single res would have used.
in parallel combination:
eq res: R=r1r2/(r1+r2)  so R is less than individual res r1 or the value of current in paallel will be greater.
and current in the ckt: Ip=V(r1+r2)/r1r2........(2)
so Ip/Is  =(r1+r2)2/r1r2 sh always be greater than 1, ie current in parallel is greater than current in series.

Answered by  | 24th Sep, 2012, 10:36: AM

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