Question

Three resistors with values of 60.0 Ω, 30.0 Ω, 

and 20.0 Ω, respectively, are connected in 

series to a 110.0 V battery of negligible 

internal resistance. Draw a circuit diagram 

and find the (a) equivalent resistance of 

the combined resistors, (b) current flowing 

through each resistor, (c) voltage drop 

across each resistor, and (d) the power 

dissipated by each resistor.

Answer 1

Equivalent resistance of combined resistors is:

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

Сurrent flowing through each resistor:

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

Voltage drop across each resistor:

$U_{R1}=IR_1=1\cdot 60=60\space V,\newline U_{R2}=IR_2=1\cdot 30=30\space V,\newline U_{R3}=IR_3=1\cdot 20=20\space V.$

The power dissipated by each resistor:

$P_{R_1}=I^2R_1=1^2\cdot 60=60\space W,\newline P_{R_2}=I^2R_2=1^2\cdot 30=30\space W,\newline P_{R_3}=I^2R_3=1^2\cdot 20=20\space W.$