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  • For the circuit shown in Fig., the equivalent resistance between and is <img alt="" src="https://cdn.entrance360.com/media/uploads/2023/08/18/f52_q34.PNG" style="height:89px; width:148px"

For the circuit shown in Fig., the equivalent resistance between and is

Option: 1

\mathrm{\frac{12}{11}r}


Option: 2

\mathrm{\frac{13}{11}r}


Option: 3

\mathrm{\frac{14}{11}r}


Option: 4

\mathrm{\frac{15}{11}r}


Answers (1)

best_answer

At junction E or F :

\mathrm{I_1+I_3+I_4=I_2-I_3 \Rightarrow I_1-I_2+2 I_3+I_4=0 }

Loop AFBA or loop ECDE

\mathrm{r I_1=r I_2+r I_3 \Rightarrow I_1=I_2+I_3}

Loop B E F B or loop EDFE:

\mathrm{r\left(I_2-I_3\right)+r I_4=r I_3 \Rightarrow I_2-2 I_3+I_4=0 }

Loop AFDCMA or ABECMA:

\mathrm{V=r I_1+r\left(I_2-I_3\right)+r I_2 \Rightarrow V=r I_1+2 r I_2-r I_3}

Solve to get \mathrm{I_1=\frac{2 V}{5 r}, I_2=\frac{V }{ r}}

\mathrm{R_{\mathrm{eq}}=\frac{V}{I}=\frac{V}{I_1+I_2}=\frac{V}{2 V / 5 r+V / 3 r}=\frac{15 r}{11} }

Posted by

Nehul

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