1.

Consider the circuit shown in the figure. The value of the resistance X for which the thermal power generated in it is practically independent of small variation of its resistance is

Answer»

`X=R`
`X=(R )/(3)`
`X=(R )/(2)`
`x=2R`

Solution :thetotalresistanceof thecircuit
` R. = R+(RX )/( R+X )`
the currentof thecircuitwill be
` (##MTG_WB_JEE_PHY_C13_E02_010_S01.png" width="80%">
`I = (V )/(R )=(E )/( R+(RX )/( R +X))`
thepotentialdropacrossthe resistanceR and Xwill be
` i.e,V_(RX )= i R_(XR ) = ((E )/( R+(RX )/( R+X ) )) ((RX )/(R xx X )) = (EX )/( R + 2X )`
the powergeneratedacrossresistanceX ,
` P_X =(V^2 RX )/( X )= (E^2X )/( (R +2X)^2)`
differtiatingwith RESPECT TOX
`(dP_X )/( dX ) = (E^2(R - 2X ))/((R + 2X)^3 ):.(dP_X )/( dx)= 0`
` thereforeR-2X=0orX =(R )/(2 )`s


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