The emf of a cell is balanced at 52 cm length of the potentiometer wire. If 5 ohm resistance is inserted fromthe resistance box connected with the cell, then balancing length obtained is 40 cm The internal resistance ofthe cell will be About the author Claire
Answer: 1.5 Ω Given, R=5Ω l 1 =52cm,l 2 =40cm To find, Internal resistance of cell, r=? r=R( l 2 l 1 −1)=5( 40 52 −1)=5( 40 52−40 )=5( 40 12 )=5( 10 3 )= 10 15 =1.5Ω ⇒r=1.5Ω Therefore, the internal resistance of the cell is 1.5Ω Reply
Answer:
1.5 Ω
Given,
R=5Ω
l
1
=52cm,l
2
=40cm
To find,
Internal resistance of cell, r=?
r=R(
l
2
l
1
−1)=5(
40
52
−1)=5(
40
52−40
)=5(
40
12
)=5(
10
3
)=
10
15
=1.5Ω
⇒r=1.5Ω
Therefore, the internal resistance of the cell is 1.5Ω