Physics, asked by chutkiaki001, 1 year ago

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 saka82411
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!!!

chutkiaki001: thanks
saka82411: my pleasure {chutki}
Answered by Samanvayasl
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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