the two resistors in parallel have a combined resistance of 30R/(30+R)
So the current in the circuit is 120/(6 + 30R/(30+R)) = 10(30+R) / 3(5+R)
Now just figure out what portion of that current goes through R, and set the power (I^2 R) equal to that dissipated by the 6Ω resistor.
Then solve for R.
A 30Ω resistor is connected in parallel with a variable resistor R. the parallel combination is then connected in
series with a 6Ω resistor and across a 120V source. Find the minimum value of R if power taken by R is equal to
the power taken by the 6Ω resistor.
1 answer