## 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 $R_5$ ## 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:

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

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

$=(9 V).(0.67)=6 V$ 