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  • In the circuit as shown in figure, the current in ammeter is:

In the circuit as shown in figure, the current in ammeter is:

 

Option: 1

\frac{42}{32}\mathrm{A}


Option: 2

\frac{27}{32}\mathrm{A}


Option: 3

\frac{15}{32}\mathrm{A}


Option: 4

\frac{32}{15}\mathrm{A}


Answers (1)

best_answer

The equivalent emf of the two cells in parallel circuit,
\varepsilon_{\mathrm{eq}}=\frac{\varepsilon_1 \mathrm{r}_2+\varepsilon_2 \mathrm{r}_1}{\mathrm{r}_1+\mathrm{r}_2}=\frac{3 \times 1+6 \times 2}{2+1}=5 \mathrm{~V}
The effective internal resistance of two cells in parallel circuit,
\mathrm{r}_{\mathrm{eq}}=\frac{\mathrm{r}_1 \mathrm{r}_2}{\mathrm{r}_1+\mathrm{r}_2}=\frac{2 \times 1}{2+1}=\frac{2}{3} \Omega
The equivalent circuit will be as shown in figure.

                                                   
Current in ammeter, \mathrm{I}=\frac{5 \mathrm{~V}}{10 \Omega+\frac{2}{3} \Omega}=\frac{15}{32} \mathrm{~A}

Hence option 3 is correct.

Posted by

manish painkra

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