1. Problem statement: We have a circuit with a voltage source $V_1=20$ V in parallel with a resistor $R_1=10\,\Omega$, and that parallel pair is in series with $R_2=5\,\Omega$, $R_3=20\,\Omega$, and a second voltage source $V_2=100$ V. 2. Unknowns and sign conventions: Let the node at the left of the series chain be the reference node where the parallel elements meet, and define the series current $I_S$ as the current flowing from that left node, through $R_2$ then $R_3$ toward $V_2$ (positive direction left-to-right as drawn). Let $I_{R1}$ be the current through $R_1$ from left node to right node of the parallel branch, and $I_{V1}$ be the current through the source $V_1$ from left node to right node of the parallel branch with the same sign convention. 3. KCL at the parallel node: The series current equals the sum of the two branch currents leaving the left node into the parallel pair, so the governing KCL equation is $$I_S=I_{V1}+I_{R1}.$$ 4. KVL around the loop through $V_1$ and the series elements: Travelling around the loop and summing drops, we get the KVL form $$-V_1+I_SR_2+I_SR_3+V_2=0.$$ Rearrange to isolate $I_S$: $$I_S(R_2+R_3)=V_1-V_2.$$ 5. Solve for the series current $I_S$ by substitution of numeric values and show intermediate simplification with cancellation: $$I_S=\frac{V_1-V_2}{R_2+R_3}$$ $$I_S=\frac{20-100}{5+20}$$ $$I_S=\frac{-80}{25}$$ $$I_S=\frac{\cancel{-80}}{\cancel{25}}= -\frac{16}{5} = -3.2\ \text{A}.$$ (Interpretation: the negative sign means the actual current flows opposite the assumed left-to-right direction; its magnitude is 3.2 A flowing right-to-left.) 6. Current through $R_1$ by Ohm's law and cancellation when simplifying: Because $R_1$ is in parallel with $V_1$, the voltage across $R_1$ equals $V_1$, so $$I_{R1}=\frac{V_1}{R_1}=rac{20}{10}=rac{\cancel{20}}{\cancel{10}}=2\ \text{A}.$$ (The positive sign means $I_{R1}$ flows from the left node to the right node as assumed.) 7. Current through $V_1$ from KCL: Use $I_S=I_{V1}+I_{R1}$ and solve for $I_{V1}$. $$I_{V1}=I_S-I_{R1}$$ Substitute the values: $$I_{V1}=-3.2-2=-5.2\ \text{A}.$$ (The negative sign means the actual current through $V_1$ flows opposite the assumed left-to-right direction, i.e., 5.2 A from right node to left node through $V_1$.) 8. Voltages across resistors $R_2$ and $R_3$: Compute using $V=IR$ with the sign consistent with the assumed $I_S$ direction. $$V_{R2}=I_SR_2=(-3.2)(5)=-16\ \text{V}.$$ $$V_{R3}=I_SR_3=(-3.2)(20)=-64\ \text{V}.$$ (These negative signs mean the potential drop in the assumed left-to-right direction is negative, consistent with the actual current being right-to-left.) 9. Voltages across the parallel elements and the sources: The resistor $R_1$ has the same voltage as $V_1$, so $$V_{R1}=V_1=20\ \text{V}.$$ The source voltages are given as $$V_1=20\ \text{V},\quad V_2=100\ \text{V}.$$ 10. Final summarized numeric results with directions interpreted physically: - Series current: $I_S=-3.2$ A (magnitude 3.2 A flowing right-to-left). - Current through $R_1$: $I_{R1}=2$ A (left-to-right). - Current through $V_1$: $I_{V1}=-5.2$ A (magnitude 5.2 A flowing right-to-left through $V_1$). - Voltage across $R_1$: $V_{R1}=20$ V (left node higher than right node as given by $V_1$ polarity). - Voltage across $R_2$: $V_{R2}=-16$ V (drop opposite assumed direction; magnitude 16 V). - Voltage across $R_3$: $V_{R3}=-64$ V (drop opposite assumed direction; magnitude 64 V). - Source voltages: $V_1=20$ V, $V_2=100$ V. All results satisfy KCL and KVL (check: $V_{R2}+V_{R3}+V_2-V_1=-16-64+100-20=0$ confirming KVL).