Four resistances 10 Omega,5 Omega,7 Omega | vedclass

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(B) $1$. The resistances are connected as follows: $AB = 10 \ \Omega$,$BC = 5 \ \Omega$,$CD = 7 \ \Omega$,$DA = 3 \ \Omega$,and diagonal $AC = 10 \ \O

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$5$

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(B) $1$. The resistances are connected as follows: $AB = 10 \ \Omega$,$BC = 5 \ \Omega$,$CD = 7 \ \Omega$,$DA = 3 \ \Omega$,and diagonal $AC = 10 \ \Omega$.$2$. When calculating the equivalent resistance between $A$ and $B$,the path $ADC$ is in series. The resistance of path $ADC$ is $R_{ADC} = R_{AD} + R_{DC} = 3 \ \Omega + 7 \ \Omega = 10 \ \Omega$.$3$. This path $ADC$ is in parallel with the diagonal resistance $AC = 10 \ \Omega$. The equivalent resistance of this parallel combination is $R_{AC}' = \frac{10 	imes 10}{10 + 10} = 5 \ \Omega$.$4$. Now,this $R_{AC}'$ is in series with the resistance $BC = 5 \ \Omega$. So,the resistance of the branch $ABC$ is $R_{ABC} = R_{AC}' + R_{BC} = 5 \ \Omega + 5 \ \Omega = 10 \ \Omega$.$5$. Finally,this branch $ABC$ is in parallel with the resistance $AB = 10 \ \Omega$. The equivalent resistance between $A$ and $B$ is $R_{eq} = \frac{10 	imes 10}{10 + 10} = 5 \ \Omega$.

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