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?
Answers
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