Answer to Question #298158 in Electric Circuits for cols

Question

Three resistors with values of 60.0Ω, 30.0Ω, 20.0Ω, 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

The diagram for the circuit is

The total resistance will be

$R_T=R_1+R_2+R_3 \\ R_T=(60.0+30.0+20.0)\,Ω=110.0\,Ω$

Then, the equivalent current intensity will be equal to the current flowing on each resistance and that will help us to calculate the voltage drop in volts and power dissipated in watts:

$\\ \therefore I_1=I_2=I_3=I_T \\ I_T= \frac{ V_T}{R_T}= \frac{ 110.0\,V}{110.0\,Ω}= 1.0\,A$

$V_1=R_1I_1=(60\,Ω)(1\,A)=60\,V \\ V_2=R_2I_2=(30\,Ω)(1\,A)=30\,V \\ V_3=R_3I_3=(20\,Ω)(1\,A)=20\,V$

$P_1=V_1I_1=(60\,V)(1\,A)=60\,W \\ P_2=V_2I_2=(30\,V)(1\,A)=30\,W \\ P_3=V_3I_3=(20\,V)(1\,A)=20\,W$

In conclusion, (a) the equivalent resistance of the whole circuit is 110.0 Ω, (b) the current flowing through each resistor is 1.0 A, (c) the voltage drop for the resistors is V₁=60.0 V, V₂=30.0 V, V₃=20.0 V, and (d) the power dissipated on each resistor is 60.0 W, 30.0 W and 20.0 W, respectively.