Chapter 28141P28.25Label the currents in the branches as shown in the first figure.Reduce the circuit by combining the two parallel resistors as shownin the second figure.Apply Kirchhoff’s loop rule to both loops in Figure (b) to obtain:2 711 7125012..RIaf+=and1 713 71500.With R=1000 Ω, simultaneous solution of these equations yields:I110 0=. mAandI2130 0=.From Figure (b),VV I IRca−=+=171240bg. V.Thus, from Figure (a),IVVR4424060 0=−==V4 000 mAΩFinally, applying Kirchhoff’s point rule at point ain Figure (a)gives:II I=−=−=+4160 010 050 0.mAmAmA,orIae=50 0. mA from point to point .(a)(b)FIG. P28.25P28.26Name the currents as shown in the figure to the right. Then wxzy++=. Loopequations are−−+=−++−=+−−+=20040 0 80 0080 040 0 360
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