73. (a) We reduce the parallel pair of identical 2 . 0 Ω resistors (on the right side) to R0 =1 . 0Ω ,andwe reduce the series pair of identical 2 . 0 Ω resistors (on the upper left side) toR 0 =4 . 0Ω . W ith R denoting the 2 . 0 Ω resistor at the bottom (between V 2 and V 1 ), we now have three resistors in series which are equivalent toR + R0 + R 0 =7 . 0Ω across which the voltage is 7.0 V (by the loop rule, this is 12 V − 5 .0 V), implying that the current is 1.0 A (clockwise). Thus, the voltage across R 0 is (1 . 0 A)(1 . 0Ω
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