Question #25dff

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Question #25dff

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Answer 1 · 2017-07-01T22:30:26

$R_{AB} = \frac{4 R}{3}$ $R_("AB")=(4R)/3$
$You can also solve it using Delta-Star conversion.$ $"You can also solve it using Delta-Star conversion."$

Explanation:

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$R_{1} and R_{2} are connected in series. R_{8} = R_{1} + R_{2} = 2 R$ $R_1" and " R_2 " are connected in series."R_8=R_1+R_2=2R$

$R_{3} and R_{4} are connected in series. R_{9} = R_{3} + R_{4} = 2 R$ $R_3" and " R_4 " are connected in series."R_9=R_3+R_4=2R$

$\frac{R_{8}}{R_{6}} = \frac{2 R}{R} = 2$ $(R_8)/(R_6)=(2R)/R=2$

$\frac{R_{9}}{R_{7}} = \frac{2 R}{R} = 2$ $(R_9)/(R_7)=(2R)/R=2$

$\frac{R_{8}}{R_{9}} = \frac{2 R}{2 R} = 1$ $(R_8)/(R_9)=(2R)/(2R)=1$

$\frac{R_{6}}{R_{7}} = \frac{R}{R} = 1$ $(R_6)/(R_7)=R/R=1$

Since the resistor ratios in the contacts are equal, no current flows through R5.

$R_{8} and R_{9} are connected in series. R_{x} = R_{8} + R_{9} = 4 R$ $R_8" and " R_9 " are connected in series."R_x=R_8+R_9=4R$

$R_{6} and R_{7} are connected in series. R_{y} = R_{6} + R_{7} = 2 R$ $R_6" and " R_7 " are connected in series."R_y=R_6+R_7=2R$

$R_{x} and R_{y} are connected in parallel. R_{AB} = \frac{R_{x} \cdot R_{y}}{R_{x} + R_{y}}$ $R_x" and " R_y " are connected in parallel. "R_("AB")=(R_x*R_y)/(R_x+R_y)$

$R_{AB} = \frac{4 R \cdot 2 R}{4 R + 2 R}$ $R_("AB")=(4R*2R)/(4R+2R)$

$R_{AB} = \frac{8 R^{2}}{6 R}$ $R_("AB")=(8R^2)/(6R)$

$R_{AB} = \frac{4 R}{3}$ $R_("AB")=(4R)/3$

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Question #25dff

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