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A battery with = 8.00 V and no internal resistance supplies current to the circuit shown in the figure below. When the double

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E = 8.0V Io = 1.05mA = 1.054103A Ta - 1.24n = 1-94x1024 Tb 7.05 mA = 3.05 X10A -) open positron :- 9 - Req = R + Rq TR3 Rob Pw submit ③ in R, TR. + R₂ = 7619 3902.44 +R2 : 7619 Rg = 7619 - 3902.44 R3= 3716.56 ar Rz 3717 * = 3.717X103 TR₂ Ž 3.717/ka -

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