A star network consists of a central node O connected to three points A , B , and C . Point A is maintained at a potential of 24 V through a branch of resistance 12 , and point B is maintained at 6 V through a branch of resistance 6 . Point C is connected to ground ( 0 V ) through a variable resistor R . The value of R for which the power dissipated in it is maximum will be

Step-by-step solution

Let the potential at the central node O be V_O . Applying Kirchhoff's Current Law at node O : 24 - V_O 12 + 6 - V_O 6 = V_O R 24 - V_O + 12 - 2V_O 12 = V_O R 36 - 3V_O 12 = V_O R 3 - V_O 4 = V_O R V_O ( 1 4 + 1 R ) = 3 V_O = 12R R + 4 The power dissipated in resistor R is: P = V_O^2 R = 144R (R + 4)^2 To maximize power, we can minimize the denominator divided by R : (R + 4)^2 R = R + 16 R + 8 By AM-GM inequality or differentiation, R + 16 R is minimum when R = 16 R , which gives R = 4 . Alternatively, using the Max