Question

A galvanometer has an external resistance of 45.0 omega and deflects full scale for a current of 1.50mA. Show how to convert the galvanometer into an ammeter that deflects full scale for an utter of 3.00 A.

Part B: a car battery has an emf of 12.5v and an internal resistance of 0.025 omega. Find the terminal potential for the battery when it is producing 160A of current

Answer 1

A.

We neet a shunt resistor, so that the ammeter has a maximum deflection when the current down the shunt is I = 3A. The rest of the current goes down the parrallel path with resistor R.
   so, (3 A) (R) = (1.50 x 10 ^-3) (r) where r = 45 ohms.
   R = (1.50*10^-3 / 3) x 45 ohms = 22.5 x 10 ^-3ohms.

B

The terminal potential for the battery is given by

                                V=e-ir =12.5 -160*0.025=12.5-4 =8.5V