a cell supplies a current of 2A when it is connected to a 5 ohms resistance and supplies a current of 1.2 A, if it is connected to a resistance 9 ohms. find the e.m.f. and internal resistance of the cell.
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Answers
Answered by
40
Hii friend,
Given,
I=>2A
R=> 5ohm
emf (V) => 2×5. {By ohms law ,V=> IR}
Emf (V) => 10 V
Supplied emf (E) => ?
E=> IR
E=> 1.2 ×9
E => 10.8 V
Formula:-
Internal resistance (r) => (E-V/V)R
r =>(10.8-10/10)5
r => 0.4 ohm.
Hope this helps you a little!!!
Given,
I=>2A
R=> 5ohm
emf (V) => 2×5. {By ohms law ,V=> IR}
Emf (V) => 10 V
Supplied emf (E) => ?
E=> IR
E=> 1.2 ×9
E => 10.8 V
Formula:-
Internal resistance (r) => (E-V/V)R
r =>(10.8-10/10)5
r => 0.4 ohm.
Hope this helps you a little!!!
Answered by
0
Answer:
Internal Resistance of the cell is 0.8 V
Explanation:
EMF of the cell
E(V)₁ = IR
I = Current of the circuit when connected to 5 ohm resistor
R = Resistance connected
E(V) = IR
= 2 ˣ 5 = 10 V
EMF of the circuit
E(V)₂ = IR
I = Current when connected to 9 ohm resistor
R= resistance of resistor
E(V)₂ = 1.2 ˣ 9
= 10.8 V
The internal resistance will be the difference between the EMF while being connected to two different resistors.
Internal Resistance = E(V)₂ - E(V)₁ = 10.8 - 10 = 0.8 V
Internal Resistance is the resistance of the cell when not coonected to a resistor.
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