three resistors are wired in series with a 22v battery. The resistances are 22.5 omega 33.6...
Homework Answers
Voltage (v) = 22v
Total resistance in series is Req = 22.5 + 33.6 + 9.9 = 66 omega
So current in the circuit is ( I )= voltage / total resistance = 22/66 = 0.333A
So voltage across the resistor 9.9 omega is (v) = current * resistance = 0.333 * 9.9
Voltage = 3.2967 volt = 3.3 volt
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