Question #317569

Three resistors with values of 60.0Ω, 30.0 Ω, and 20.0 Ω, respectively, are


connected in series to a 110.0 V battery of negligible internal resistance. (a) Draw a circuit diagram and find the (b) equivalent resistance of the combined resistors, (c) current flowing through each resistor, (d) voltage drop across each resistor.


Expert's answer


Equivalent resistance of combined resistors is:


R = R1 + R2 + R3 = 60 + 30 + 20 = 110 Ohm.


Сurrent flowing through each resistor:


I=ER=110110=1 A.I=\frac{E}{R}=\frac{110}{110}=1\space A.


Voltage drop across each resistor:


UR1=IR1=160=60 V,UR2=IR2=130=30 V,UR3=IR3=120=20 V.U_{R1}=IR_1=1\cdot 60=60\space V,\newline U_{R2}=IR_2=1\cdot 30=30\space V,\newline U_{R3}=IR_3=1\cdot 20=20\space V.





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