Direct-Current Circuits133Substitution of equation (4) into (3) yields 11140310450II+=or 2113104501400II−Solving this quadratic equation gives two possible values for the current . These are . Then, from 1I1A and 0.452=11.0A=2120 W=RI, we find two possible values for the resistance R. These are 20 or 98RR=Ω=Ω18.57When connected in series, the equivalent resistance is 12eqnRRRRnR=++⋅⋅⋅. Thus, the current is ( )( )seqIVR V
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