(a) Two cells of different emfs and internal resistances are connected in parallel with one another. Derive the expression for the equivalent emf and equivalent internal resistance of the combination.

(b) Two identical cells of emf 1·5 V and internal resistance r are each connected in parallel providing a supply to an external circuit consisting of two resistances of 17\; \Omega each joined in parallel. A very high resistance voltmeter reads the terminal voltage of the cell to be 1·4 V. Calculate the internal resistance of each cell.

 

 

 

 
 
 
 
 

Answers (1)
S safeer

(a) Let the total amount (I)=I_{1}+I_{2} ______(i)

     

\because V=E_{1}-I_{1}r_{1}

\because I_{1}=\frac{E_{1}-V}{r_{1}} and

\because V=E_{2}-I_{2}r_{2}

\because I_{2}=\frac{E_{2}-V}{r_{2}}

On putting the values of I_{1} and I_{2} in equation (i) ,we get 

I=\frac{E_{1}-V}{r_{1}}+\frac{E_{1}-V}{r_{2}}

I=\left [ \frac{E_{1}}{r_{1}}+\frac{E_{2}}{r_{2}} \right ]-V\left [ \frac{1}{r_{1}}+\frac{1}{r_{2}} \right ]

V=\frac{E_{1}r_{2}+E_{2}r_{1}}{r_{1}+r_{2}}-\frac{Ir_{1}r_{2}}{r_{1}+r_{2}} 

Now, comparing with;-

V=E_{eq}-Ir_{eq}

Then, we get the values of E_{eq} and r_{eq} 

E_{eq}=\frac{E_{1}r_{2}+E_{2}r_{1}}{r_{1}+r_{2}}

r_{eq}=\frac{r_{1}r_{2}}{r_{1}+r_{2}}

(b) 

        V=1.4 \; Volt and emf (E_{eq})=1.5\; V

   r_{eq}=\frac{17}{2}=8.5\; \Omega

To calculate the internal resistance 

We have,

I=\frac{V}{r_{eq}}

On putting the values, we get


I=\frac{1.4}{8.5}\; Ampere

Now, V=E_{eq}-Ir_{eq}

 

So, 1.4=1.5-\frac{1.4}{8.5}\times \frac{r}{2}\; \; \; \; \left ( \because r_{eq}=\frac{r}{2} \right )

Hence, r=1.21\; \Omega is the required internal resistance.

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