Consider the following circuit shown below. A | vedclass

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(A) First,we distribute the current in the circuit as shown in the diagram.The current distribution must follow Kirchhoff's junction rule.Now,from the

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$8$

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(A) First,we distribute the current in the circuit as shown in the diagram.The current distribution must follow Kirchhoff's junction rule.Now,from the closed loops marked $1, 2, 3$ and $4$,we have the following set of equations by applying Kirchhoff's loop rule:$I_1 = I_2 + I_3 \quad \dots(i)$$I_3 = I_2 + I_4 \quad \dots(ii)$$I_4 = I_2 - I_4 + I_5$$\Rightarrow 2 I_4 = I_2 + I_5 \quad \dots(iii)$$I_5 = 2(I_2 - I_4 - I_5)$$\Rightarrow I_5 = 2 I_2 - 2 I_4 - 2 I_5 \quad \dots(iv)$$3 I_5 = 2 I_2 - 2 I_4 \quad \dots(v)$From Eqs. $(iii)$ and $(v)$,we have:$3 I_5 = 2 I_2 - (I_2 + I_5)$$\Rightarrow 4 I_5 = I_2 \quad \dots(vi)$From Eqs. $(iii)$ and $(vi)$,we have:$2 I_4 = 4 I_5 + I_5 \Rightarrow I_4 = \frac{5}{2} I_5 \quad \dots(vii)$From Eqs. $(ii), (vi)$ and $(vii)$,we have:$I_3 = 4 I_5 + \frac{5}{2} I_5 = \frac{13}{2} I_5 \quad \dots(viii)$Now,the marked currents $I$ and $I^{\prime}$ in the given circuit are:$I^{\prime} = (I_2 - I_4 - I_5) = (4 I_5 - \frac{5}{2} I_5 - I_5)$$= (\frac{8 - 5 - 2}{2}) I_5 = \frac{I_5}{2} \quad \dots(ix)$And $I = I_2 = 4 I_5$Hence,the ratio of $I / I^{\prime} = (4 I_5) / (I_5 / 2) = 8$.

Consider the following circuit shown below. All the resistors are identical. The ratio of $I / I^{\prime}$ is

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