code class="asciimath">J_(eq)=J_(1)+((Z_(1))/(Z_(2)))^(2)J_(2) $$J_{eq} = J_1 + \left(\frac{Z_1}{Z_2}\right)^2 J_2$$
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$J_{0}(x)-J_{2}(x)=2(d / d x) J_{1}(x)$
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$J_{1}(x)+J_{3}(x)=(4 / x) J_{2}(x)$
(a) The currents are as shown. From Ohm's law applied between 1 and 7 via 1487 (say) (b) Between 1 and 2 from the loop 14321 , $I_{1} R=2 I_{2} R+I_{3} R$ or, $I_{1}=I_{3}+2 I_{2}$ From the loop 48734 , $\left(I_{2}-I_{3}\right) R+2\left(I_{2}-I_{3}\right) R+\left(I_{2}-I_{3}\right) R=I_{3} R$ or, $4\left(I_{2}-I_{3}\right)=I_{3}$ or $I_{3}=\frac{4}{5} I_{2}$ so $I_{1}=\frac{14}{5} I_{2}$ Then, $\left(I_{1}+2 I_{2}\right) R_{e q}=\frac{24}{5} I_{2} R_{\mathrm{eq}}=I_{1} R=\frac{14}{5} I_{2} R$ or $R_{e q}=\frac{7}{12} R$ (c) Between 1 and 3 From the loop 15621 $$ \begin{aligned} &\qquad I_{2} R=I_{1} R+\frac{I_{1}}{2} R \text { or, } I_{2}=3 \frac{I_{1}}{2} \\ &\text { Then, } \begin{array}{c} \left(I_{1}+2 I_{2}\right) R_{e q}=4 I_{1} R_{e q} \\ & =I_{2} R+I_{2} R=3 I_{1} R \\ \text { Hence, } \quad R_{e q}=\frac{3}{4} R \end{array} \end{aligned} $$
Electrodynamics
Electric Current
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