If voltage is applied between terminals 1 and 2 when terminals 3 and 4 are open the power liberated is `P_(1)=40W` and when terminals 3 and 4 are connected, the power liberated is `P_(2)=80W`. If the same source is connected to the terminals 3 and 4, the power liberated in the circuit when terminals 1 and 2 are open in `P_(3)=20W`. Determine the power `P_(4)` consumed in the circuit when the terminals 1 and 2 are connected and the same voltage is applied between the terminals 3 and 4.

Answer

40w

Explanation

When 1 and 2 are connected to voltage suppy V and 3 and are open
`P_(1)=(R_(1)+R_(3))((V)/(R_(1)+R_(3)))^(2)=(V^(2))/(R_(1)+R_(3))`
When 3 and 4 are closed and 1 and 2 are connected to `V`
`R_(eq)=R_(1)+(R_(2)R_(3))/(R_(2)+R_(3))becauseP_(2)=(V^(2)(R_(2)+R_(3)))/(R_(1)R_(2)+R_(2)R_(3)+R_(3)R_(1))`
When 3 amd 4 are connected to V and 1 and are open `P_(3)=(V^(2))/(R_(2)+R_(3))`
When 3 and 4 are connected to V and 1 and 2 are closed:
`R_(eq)=R_(2)+(R_(3)R_(1))/(R_(3)+R_(1))=(R_(1)R_(2)+R_(2)R_(3)+R_(3)R_(1))/(R_(3)+R_(1))`
`thereforeP_(4)=(V^(2)(R_(3)+R_(1)))/(R_(1)R_(2)+R_(2)R_(3)+R_(3)R_(1))`
Now, `(P_(1))/(P_(3))=(R_(2)+R_(3))/(R_(1)+R_(3))` and `(P_(2))/(P_(4))=(R_(2)+R_(3))/(R_(1)+R_(3))`
`implies(P_(1))/(P_(3))=(P_(2))/(P_(4))` or `P_(4)=(P_(2))/(P_(1))xxP_(3)=(80xx20)/(40)=40W`