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Three resistances are connected to form a *T*-shape as shown in the figure. Then the current in the 4Ω resistor is

(a) 0.93 A (b) 1.42 A (c) 2.5 A (d) 1.57 A

We will solve this using Kirchhoff's first law at junction B.

Let current throug 2ohm is I and 8ohm I

_{1}and 4ohm I

_{2}.

Then

$I={I}_{1}+{I}_{2}\phantom{\rule{0ex}{0ex}}{I}_{1}=I-{I}_{2}\phantom{\rule{0ex}{0ex}}$

By applying voltage law

Point A is at +2V, Point B is junction point. Point C is at -4V and D at -8V.

${V}_{AB}=\sum ir-\sum E\phantom{\rule{0ex}{0ex}}-2-4=-2I-8{I}_{1}\phantom{\rule{0ex}{0ex}}6=2I+8{I}_{1}\phantom{\rule{0ex}{0ex}}6=2I+8(I-{I}_{2})\phantom{\rule{0ex}{0ex}}6=10I-8{I}_{2}.....Equation-1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}-2+8=2I+4{I}_{2}\phantom{\rule{0ex}{0ex}}6=2I+4{I}_{2}...Equation-2\phantom{\rule{0ex}{0ex}}Solving1and2wegetI2=1.57A\phantom{\rule{0ex}{0ex}}$

Regards

Vijaya

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