Chapter 2

Q. 2.4

Determine the following for the DC circuit shown below:
a. Equivalent resistance for the entire circuit, if R_1 = 5  Ω, R_2 = R_3 = 10  Ω , and R_4 = R_5 = 20  Ω
b. Current flowing through resistor R_1
c. Voltage across resistor R_5

2.4.1

Step-by-Step

Verified Solution

a. Combination of R_2 and R_4 = R_{2,4} = R_2 + R_4 = 10  Ω + 20  Ω = 30  Ω Combination of R_3 and R_5 = R_{3,5} = R_3 + R_5 = 10  Ω + 20  Ω = 30  Ω Combination of R_{2,4} and R_{3,5} =

R_{2-5}=\frac{(30  Ω).(30  Ω)}{(30  Ω+30  Ω)}=\frac{900}{60}=15  Ω

 

R_{eq}=R_1+R_{2-5}=5  Ω+15  Ω=20  Ω

b. Current through R_1 would be the same as the current through the 12 V supply:

I=\frac{V}{R_{eq}}=\frac{12  V}{20  Ω}=0.6  A

c. One method for determining V_{R5}, voltage across R_5, is to first calculate V_{R2–5}, the voltage across the combined resistance of resistances R_2, R_3, R_4, and R_5. Then, by applying voltage division, calculate V_{R5} as follows:
According to Ohm’s law,

V_{R2–5}=I.(R_{2–5})=(0.6  A).(15  Ω)=9  V

Then, by applying the voltage division rule:

V_{R5}=(9  V).\left\lgroup\frac{R_5}{R_5+R_3}\right\rgroup

 

=(9  V).\left\lgroup\frac{20  Ω}{20  Ω+10  Ω}\right\rgroup

 

=(9  V).(0.67)=6  V
2.4.2