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  • Seven resistors, each of  resistance are connected as shown in figure. What is the effective resistance

Seven resistors, each of 1 \Omega resistance are connected as shown in figure. What is the effective resistance between A and B?

Option: 1

\mathrm{\frac{4}{7}\Omega}


Option: 2

\mathrm{7\: \Omega}


Option: 3

\mathrm{\frac{8}{7}\Omega}


Option: 4

\mathrm{\frac{3}{2}\Omega}


Answers (1)

best_answer

Let a cell of emf \varepsilon be connected between A and B. The currents through the various arms will be as shown in the figure.

Applying Kirchhoff’s loop law in closed loop ACEA, we get

\begin{aligned} & -\mathrm{I}_1-\mathrm{I}_2+\left(\mathrm{I}-\mathrm{I}_1\right)=0 \\ \\& \mathrm{I}=2 \mathrm{I}_1+\mathrm{I}_2\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \text{(i)} \end{aligned}

Again applying Kirchhoff’s loop law in closed CEDC, we get

-\mathrm{I}_2-\mathrm{I}_2+\left(\mathrm{I}_1-\mathrm{I}_2\right)=0 \quad \text { or } \quad 3 \mathrm{I}_2=\mathrm{I}_1

Putting this value in (i) , we get

\mathrm{I}=2 \mathrm{I}_1+\frac{\mathrm{I}_1}{3} \quad \text { or } \quad \mathrm{I}_1=\left(\frac{3}{7}\right) \mathrm{I}

Again applying Kirchhoff’s loop law in closed loop AEBA, we get

\begin{aligned} & \mathrm{ -\left(I-I_1\right)-\left(I-I_1\right)+\varepsilon=0} \\ \\& \mathrm{ \varepsilon=2\left(I-I_1\right)} \\ \\& \mathrm{ \varepsilon=2\left(I-\frac{3 I}{7}\right)=\frac{8 I}{7}} \end{aligned}

If R is the effective resistance between A and B, then

\mathrm{\varepsilon = IR}

So, \mathrm{\mathrm{IR}=\frac{8 \mathrm{I}}{7}}                or               \mathrm{R}=\frac{8}{7} \Omega

Posted by

Devendra Khairwa

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