(1) 10 Ω(3) 109 ΩA resistor of 10 Ω connected across a cell of emf

1.	(1) 10 Ω(3) 109 ΩA resistor of 10 Ω connected across a cell of emf 12 V, draws the current of 1.1 A.Find the internal resistance of the cell(1) 10 Ω(3) 109 Ω(2)0.1620.91 Ω(2)0.1 Ω(4)091 Ω
Answer» From Ohm’s Law i = E / (R + r) 1.1 = 12 / (10 + r) 11 + 1.1r = 12 1.1r = 1 r = 1 / 1.1 r = 10/11 Ω ∴ Internal resistance of battery is 10/11 = 0.91 Ω

(1) 10 Ω(3) 109 ΩA resistor of 10 Ω connected across a cell of emf 12 V, draws the current of 1.1 A.Find the internal resistance of the cell(1) 10 Ω(3) 109 Ω(2)0.1620.91 Ω(2)0.1 Ω(4)091 Ω

Answer»

From Ohm’s Law

i = E / (R + r)

1.1 = 12 / (10 + r)

11 + 1.1r = 12

1.1r = 1

r = 1 / 1.1

r = 10/11 Ω

∴ Internal resistance of battery is 10/11 = 0.91 Ω