- In a potentiometer arrangement a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell, the balance point shifts to 63 cm. What is the emf of the second cell?
E1= 1.25 V
l1 = 35 cm
l2 = 63 cm
E1/E2 = l1/l2
E2 = l2E1/l1
E2 = 63 x 1.25/35
E2 = 2.25 V (Ans)
A Problem for Practice
A storage battery of emf 8 V and internal resistance 0.5 ohm is charged by 120V dc supply using a resistor 15.5 ohm in series. What is the terminal voltage of the battery during charging?
(Try to solve this problem and post your responses as comment to this post)
Hint: During charging; V = E +Ir
No comments:
Post a Comment
Do not spam. Spammers will be banned from this site