1.

A coil of 20 Ω resistance is joined in parallel with a coil of R Ω resistance. This combination is then joined in series with a piece of apparatus A, and the whole circuit connected to 100 V mains. What must be the value of R so that A shall dissipate 600 W with 10 A passing through it?

Answer»

Given :

20Ω resistance & R is connected in parallel & whole COMBINATION is connected to A such that 600 W power is dissipated from A when connected across 100V & 10 A current passes through it.

To Find :

VALUE of R

Solution:

•Power dissipated across A is 600W when 10A of current passes through it .

•Also , P = VI

where P is power dissipated

V is potential DIFFERENCE across A

I is current through A

=> 600 = V(10)

=> V = 60V

•So , potential difference across A is 60V

=> potential difference across R & 20Ω is 40V

•Since , this parallel combination is connected in series with A

=> Current through parallel combination of 20Ω & R is 10A

•By ohm's LAW

V = IR

40 = 10Req

=> Req = 4Ω

also , For parallel combination

1/Req = 1/R1 + 1/R2

1/4 = 1/20 + 1/R

1/4 = (R+20)/20R

20R = 4R + 80

16R = 80

=> R = 5Ω

•Hence , Value of R is 5 Ω


Discussion

No Comment Found

Related InterviewSolutions