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Two cells, having emfs of 10 \mathrm{~V}and 8 \mathrm{~V}respectively are connected in series with a resistance of 24\, \Omega in the external circuit. If the internal resistances of each of these cells in ohm are 200 \% of the value of their emp's respectively, the terminal potential difference across 8 \mathrm{~V}battery is

Option: 1

2.4 \mathrm{V}


Option: 2

2.8 \mathrm{V}


Option: 3

3.2 \mathrm{V}


Option: 4

3.6 \mathrm{V}


Answers (1)

best_answer

Internal resistances of each cells are

\mathrm{r_{1}=2 \times 10 \Omega=20 \Omega}
\mathrm{r_{2}=2 \times 8 \Omega=16 \Omega}
Total resistance in the circuit

\mathrm{=(24+20+16)\Omega=60 \Omega}
\mathrm{\Rightarrow \text { current }=\frac{V_{e q}}{R_{e q}}=\frac{18 \mathrm{~V}}{60}=0.3 \mathrm{~A}}
\Rightarrow Terminal potential difference 8 \mathrm{~V} battery is \mathrm{V =E-I r}
                                                                                 \mathrm{=(8-0.3 \times 16) \mathrm{V} }
                                                                                 \mathrm{=3.2 \mathrm{~V}}.                  

Posted by

sudhir kumar

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