Three resistors A,B , and C are connected in series to a 117 volt source . If RA= 64 ohms , and EB = 40 volts when the current is 0.5 amp , calculate the resistors RB and RC
−117+64(0.5)+40+Rc(0.5)=0Rc=450.5=90ohmsRb=400.5=80ohms-117 + 64(0.5) + 40 + R_c(0.5) = 0\\ R_c = \frac{45}{0.5} = 90ohms\\ R_b = \frac{40}{0.5} = 80 ohms\\−117+64(0.5)+40+Rc(0.5)=0Rc=0.545=90ohmsRb=0.540=80ohms
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