Question 1 of 7: DC Circuit – KCL/KVL and an Unknown Resistor
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-4 Electric Circuits and Power — December 2014 national exam. 3 hours, closed book (one aid sheet permitted). Any five of the seven questions constitute a complete paper; the first five as they appear in the answer book are marked. All seven questions are solved below as a complete study resource.
Reference texts: Sadiku & Alexander, Fundamentals of Electric Circuits (6th ed.) — DC circuit analysis, Thevenin/Norton equivalents, first-order transients, AC steady-state phasors and power, magnetic circuits, diode rectifiers (Ch. 2–4, 9–11, 13); Mano & Ciletti, Digital Design — combinational logic design (Ch. 2–4), cited inline for Question 7.
Question 1: DC Circuit – KCL/KVL and an Unknown Resistor (20 marks)
Given. $R_1=3\,\Omega$, $R_2=6\,\Omega$, $R_3=6\,\Omega$, $R_4=8\,\Omega$, $R_5=6\,\Omega$, $V_s=18\text{ V}$ (ideal source, directly across C–D); observed $I_5=2\text{ A}$. Node D is the common (ground) rail.
Given data
Quantity
Value
$R_1,R_2$ (both at node A, to ground)
$3\,\Omega,\ 6\,\Omega$
$R_3$ (node B to ground)
$6\,\Omega$
$R_4$ (node C to ground)
$8\,\Omega$
$R_o$ (A–B), $R_5$ (B–C)
unknown, $6\,\Omega$
$V_s$, observed $I_5$
$18\text{ V}$, $2\text{ A}$
[Figure not reproduced. See the official exam paper.]
Find. (a) KCL at A, B, C; (b) KVL for loops $R_1R_3R_o$ and $R_1V_sR_5R_o$; (c) $V_{BD}$ and $I_3$; (d) $R_o$, $I_o$, and $P_{R_o}$.
Approach. Because $V_s$ is an ideal source directly across C–D, node C is fixed at $18\text{ V}$; walk the ladder leftward with KVL/Ohm's law using the given $I_5$, then close the network at node A (where $R_1\parallel R_2$) to isolate $R_o$.
a) KCL at nodes A, B, C (currents defined per Figure 1, all flowing left along the top rail and down through the shunt resistors):
$$\text{Node A: } I_o = I_1 + I_2 \qquad \text{Node B: } I_5 = I_3 + I_o \qquad \text{Node C: } I_s = I_4 + I_5$$
These simply state that the current arriving at each node from the right splits between the downward shunt branch and the continuing leftward branch.
b) KVL for the two named loops. Loop $R_1R_3R_o$ (traversing $R_1$ up from D to A, $R_o$ from A to B, $R_3$ down from B to D):
$$-I_1R_1 - I_oR_o + I_3R_3 = 0$$
Loop $R_1V_sR_5R_o$ (traversing $R_1$, then across to C via the bottom rail and up through $V_s$, then $R_5$ from C to B, then $R_o$ from B to A):
$$-I_1R_1 + V_s - I_5R_5 - I_oR_o = 0$$
c) Solve for $V_{BD}$ and $I_3$. Since $V_s$ is ideal, $V_C=V_s=18\text{ V}$. Applying Ohm's law across $R_5$ with the observed $I_5$:
$$V_{BD}=V_C-I_5R_5 = 18-(2)(6)=\boxed{6\text{ V}}$$
Then $I_3$ follows directly since $R_3$ is the only path from B to ground:
$$I_3=\frac{V_{BD}}{R_3}=\frac{6}{6}=\boxed{1\text{ A}}$$
d) Solve for $R_o$, $I_o$, and $P_{R_o}$. KCL at B gives $I_o=I_5-I_3=2-1=1\text{ A}$. At node A, $R_1$ and $R_2$ are in parallel to ground, so
$$V_{AD}=I_o\left(\frac{R_1R_2}{R_1+R_2}\right)=(1)\left(\frac{3\times 6}{3+6}\right)=(1)(2)=2\text{ V}$$
Applying Ohm's law across $R_o$ (between B and A):
$$R_o=\frac{V_{BD}-V_{AD}}{I_o}=\frac{6-2}{1}=\boxed{4\ \Omega}$$
$$P_{R_o}=I_o^2R_o=(1)^2(4)=\boxed{4\text{ W}}$$
As a check, $I_2=V_{AD}/R_2=2/6=0.333\text{ A}$ and $I_1=I_o-I_2=0.667\text{ A}$, and $I_4=V_C/R_4=18/8=2.25\text{ A}$ so $I_s=I_5+I_4=4.25\text{ A}$; substituting all currents back into the loop equations from part (b) closes them to zero, confirming consistency.
