Question 1 of 6: Equivalent resistance and source loading

Question 1: Equivalent resistance and source loading [10 + (5+5)]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A $6\,\Omega$ resistor is connected directly across A–B (branch current $I_2$). A second branch leaves A through $8\,\Omega$ (current $I_1$) to node C, where a $20\,\Omega$ resistor and the unknown $R$ are connected in parallel between C and node D; from D a $10\,\Omega$ resistor returns to B. The specified terminal resistance is $R_{AB}=5\,\Omega$.

Find. (a) $R$ such that $R_{AB}=5\,\Omega$; (b) with $10\text{ V}$ across A–B, the current $I_R$ through $R$ and the total power delivered.

[Figure not reproduced: Figure 1 — redrawn. The $6\,\Omega$ is directly across A–B; the outer branch is $8\,\Omega$ in series with the parallel pair $20\,\Omega\,\|\,R$ (nodes C–D) in series with $10\,\Omega$. See the official exam paper.]

Approach. Reduce the outer branch to a single resistance in terms of $R$, set the parallel combination with the $6\,\Omega$ equal to $5\,\Omega$, solve for $R$; then re-energise with $10\text{ V}$ and use current division.

1. Combine the parallel pair. The $20\,\Omega$ and $R$ share nodes C and D, so $$R_{CD}=\frac{20R}{20+R}.$$
2. Series resistance of the outer branch. Adding the $8\,\Omega$ and $10\,\Omega$ in series, $$R_{\text{branch}} = 8 + \frac{20R}{20+R} + 10 = 18 + \frac{20R}{20+R}.$$
3. Impose the terminal condition. This branch is in parallel with the $6\,\Omega$: $$R_{AB}= \frac{6\,R_{\text{branch}}}{6+R_{\text{branch}}}=5.$$ Solving, $6R_{\text{branch}} = 5(6+R_{\text{branch}})\Rightarrow R_{\text{branch}}=30\,\Omega.$
4. Back out $R$. Then $18 + \dfrac{20R}{20+R}=30\Rightarrow \dfrac{20R}{20+R}=12\Rightarrow 20R = 12(20+R)\Rightarrow 8R=240.$ $$\boxed{R = 30\,\Omega}$$ As a check, $20\,\|\,30 = 12\,\Omega$, branch $=8+12+10=30\,\Omega$, and $6\,\|\,30 = 5\,\Omega.$ ✓
5. (b) Total current from the $10\text{ V}$ source. With $R_{AB}=5\,\Omega$, $$I_{\text{total}}=\frac{10}{5}=2\text{ A}.$$ The current splits between the $6\,\Omega$ ($I_2=10/6=1.667\text{ A}$) and the $30\,\Omega$ outer branch ($I_1=10/30=0.333\text{ A}$); $1.667+0.333=2\text{ A}$ ✓.
6. Current through $R$ by current division. $I_1=0.333\text{ A}$ reaches node C and develops $$V_{CD}=I_1\,(20\,\|\,R)=0.333\times12 = 4\text{ V},\qquad I_R=\frac{V_{CD}}{R}=\frac{4}{30}.$$ $$\boxed{I_R = 0.133\text{ A}=133.3\text{ mA}}$$ (The $20\,\Omega$ carries the remaining $4/20=0.200\text{ A}$; $0.133+0.200=0.333\text{ A}$ ✓.)
7. Power supplied by the source. $$P = V\,I_{\text{total}} = 10\times 2 = \boxed{20\text{ W}}.$$