Saved Bookmarks
| 1. |
A resistance of ROmega draws current from a potentiometer as shown in Fig. The potentiometer has a total resistance R_0 Omega. A voltage Vis supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer. |
|
Answer» Solution :When the sliding contact is in the middle of the potentiometer, total effective resistance `R_(eff)`of potentiometer is SUM of PARALLEL combination of `(R_0)/(2)`and R, and`(R_0)/(2)` ` therefore R_(eff) = ( (R_0 . R)/(2))/( (R_0)/(2) + R) + (R_0)/(2)= (R_0)/(2) [ ( R)/((R_0)/(2) + R) + 1 ] = (R_0 (2R + (R_0)/(2) ))/(2((R_0)/(2) + R) ) ` ` therefore `current in the circuit `I = (V)/(R_(eff)) = (V.2 ( (R_0)/(2) + R) )/(R_0 (2R + (R_0)/(2)) )` ` therefore ` VOLTAGE across R , `V_(eff) = I. ( (R_0 .R)/(2))/((R_0)/(2) + R)= (IR_0R)/(2((R_0)/(2) + R) ) = (V.2 ( (R_0)/(2) + R) )/(R_0 (2R + (R_0)/(2) ) ) xx (R_0 R)/(2 ((R_0)/(2) + R) ) = (V.R )/((2R + (R_0)/(2)) )` |
|