A 16, Omega wire is bent to form a square loo | vedclass

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(A) The total resistance of the wire is $16\, \Omega$,so each side of the square loop has a resistance of $R = 4\, \Omega$.Let the current supplied by

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$45$

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(A) The total resistance of the wire is $16\, \Omega$,so each side of the square loop has a resistance of $R = 4\, \Omega$.Let the current supplied by the battery be $I_{total} = 4i$. The current splits at the junction into $3i$ (through the side connected to the battery) and $i$ (through the remaining three sides in series,which have a total resistance of $4+4+4 = 12\, \Omega$).Applying Kirchhoff's Voltage Law $(KVL)$ to the outer loop containing the battery and the path with current $i$:$9 - (1)(4i) - (4)(i) - (4)(i) - (4)(i) = 0$$9 - 4i - 12i = 0$$16i = 9 \implies i = \frac{9}{16}\, A$.The potential drop across the diagonal of the square loop is the potential difference across the two resistors in series that form the diagonal path. Let the vertices be $A, B, C, D$ in order. If the battery is connected across $AB$,the diagonal is $AC$ or $BD$. The potential drop across the diagonal $AC$ is the sum of potential drops across the two sides $AB$ and $BC$ is not correct; rather,we calculate the potential difference between the two opposite corners.Using the circuit diagram,the potential drop across the diagonal is the voltage across the two resistors in series $(4\, \Omega + 4\, \Omega = 8\, \Omega)$ carrying current $i$.$V_{diag} = i 	imes 8\, \Omega = \frac{9}{16} 	imes 8 = 4.5\, V$.$4.5\, V = 45 	imes 10^{-1}\, V$.

$A$ $16\, \Omega$ wire is bent to form a square loop. $A$ $9\, V$ supply having an internal resistance of $1\, \Omega$ is connected across one of its sides. The potential drop across the diagonals of the square loop is $.......\, \times 10^{-1}\, V$.

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