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  •  The potential difference in the Circuit shown <img alt="\mathr

 The potential difference \mathrm{V_{A}-V_{B}} in the Circuit shown \mathrm{E_{1}=1.5 \mathrm{~V}, E_{2}=2.0 \mathrm{~V}}
\mathrm{E_{3}=2 \mathrm{~V}, R_{1}=10 \Omega, R_{2}=20 \Omega, R_{3}=30 \Omega}

Option: 1

\mathrm{\frac{5}{11} V}


Option: 2

2 \mathrm{~V}


Option: 3

\frac{10}{11} \mathrm{~V}


Option: 4

3 \mathrm{~V}


Answers (1)

best_answer

we can reduce the given circuit into one battery and one resistance Equivalent LMF of the Battery is

\mathrm{E_{e q}=\frac{\frac{E_{1}}{\gamma_{1}}+\frac{E_{2}}{\gamma_{2}}-\frac{E_{3}}{\gamma_{3}}}{\frac{1}{\gamma_{1}}+\frac{1}{\gamma_{2}}+\frac{1}{\gamma_{3}}}=\frac{\frac{1.5}{10}+\frac{2}{20}-\frac{2.5}{30}}{\frac{1}{10}+\frac{1}{20}+\frac{1}{30}}}
        \mathrm{=\frac{1 / 6}{11 / 60}=\frac{10}{11} \text { Vols }}

\mathrm{\frac{1}{R_{\text {eq }}}=\frac{1}{10}+\frac{1}{20}+\frac{1}{30}=\frac{6+3}{60}+2}
  \mathrm{R_{\text {eq }}=\frac{60}{11} \Omega \Rightarrow}                  

\mathrm{V_{A}-V_{B}=E_{e q}=\frac{10}{11} \mathrm{Volts}}

Posted by

Ritika Harsh

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