Question

Potential difference of a battery is 2.2v when it is connected across a resistance of 5 ohm, if suddenly potential difference falls to 1.8 v, its internal resistance will be

Answer 1

V1=IR1
I=V1/R1
=22/5*10
R2=V2/I
=18*5*10/22*10
=45/11
=4.9ohm

Answer 2

The internal resistance of the cell in the given circuit conditions is 1.11 ohms.

Given :

EMF ( E) of cell = 2.2V

The potential difference across R = 1.8V

External resistance i.e. R = 5 Ohm

To Find :

The internal resistance of the cell

Solution :

The potential difference across a battery when it is in an open circuit i.e. not connected is called its EMF. And when it is connected in a closed circuit, it is called potential difference i.e. V.

Thus, we get E = 2.2Volt  and V = 1.8Volt

Let the internal resistance of the battery be "r" and the external resistance given is R = 5 Ohm.

As both resistances are placed in series, the net effective resistance is :

                     $R_{total} = R_{internal} + R_{external}$

                                $= r + 5$

Thus, in a closed connected circuit, we have :  

Net Potential difference of battery = 2.2V

Net Resistance acting = r + 5 ohm

Then the current (I)  flowing through the wire is given by :

                         $V = I * R$

                         $2.2 = I * (5+r)$

                         $I = 2.2/(5+r)$

                   

Vint i.e. potential across the internal resistance as given is :

                         $V_{internal} = E - V_{external}$

                                        $= 2.2 - 1.8$

                                        $= 0.4 Volts$

Now across the internal resistance, we have :

Internal Potential difference = 0.4 Volts

Current flows the same along the circuit i.e. I

Internal resistance = r

Applying the same formula again to calculate the new resistance( r ),

                          $V_{internal} = I * R_{internal}$

                          $0.4 = ( 22/(5+r)) * r$

                          $0.4(5+r) = 22 * r$

                          $2 + 0.4r = 22r$

                          $1.8*r = 2$

                          $r = 1.11$

Hence, the internal resistance of the given battery is 1.11 ohms.    

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