A circuit has resistors R 1 = 468 Ω and R 2 = 125 Ω R1=468 Ω and R2=125 Ω , and two batteries V 1 = 12.0 V and V 2 = 3.00 V V1=12.0 V and V2=3.00 V . The variable resistor R is adjusted until the galvanometer reads 0 A 0 A . Calculate the unknown resistance R R .

0.0192A

Explanation:

Since, the reading of the galvanometer is 0 A, the voltage across the resistance R will be:

Step 1

VR = V2

VR = 3.00v

Step 2

Calculating the current through the resistance R as below,

IR = V1 - V2 /R1

IR = 12 - 3 /468

IR =0.0192A

im guessing the west of station b

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down