Question

A 24 v battery of internal resistance r=4 ohm connected to a variable resistance r. The rate of heat dissipated in the resistor is maximum when the current drawn from the battery is i. The current drawn from the battery will be ½ when r is equal to

Answer 1

Answer:

same

because of Maximum current flowing in the wire

Answer 2

Answer: R=12 ohm

Explanation:

Given:

$Emf\ of\ cell=24V\\Internal\ resistance(r)=4ohm$

Let, Variable resistance =R

• Condition for maximum heat to be dissipated :

           (r=R)

When maximum heat is dissipated in that case, the Current is calculated as follows:

Applying: $\Delta V=IR$(From ohm's law)-

$I=V/(R+r)$

$I=24/(4+4)=24/8=3A$

When the current drawn from the battery is half which is equal to 3/2 then, in that case :

$24/(R+4)=3/2\\48=3R+12\\3R=48-12=36\\R=12\Omega$