1. State the problem. We have a circuit with two ideal voltage sources and three resistors arranged so that the left top node (call it $V_a$) is at the positive terminal of V1 = 20 and the right top node (call it $V_b$) is at the positive terminal of V2 = 100. Resistor R1 = 10 is connected between $V_a$ and $V_b$ and resistors R2 = 5 and R3 = 20 are connected in series between $V_a$ and $V_b$. We must find the currents through each branch and the voltages across each element. 2. Key formulas and rules used. Ohm's law: $V=IR$. Series resistance: $R_{23}=R_2+R_3$. KCL (node currents): the sum of currents leaving a node is zero. KVL (loop voltages): the algebraic sum of voltage rises and drops around any closed loop is zero. 3. Assign node voltages and sign convention. Choose the circuit bottom as reference (0 V). Then by the ideal sources the node voltages are $V_a=20$ and $V_b=100$. We take currents positive from $V_a$ toward $V_b$ for the branch definitions; a negative result means actual flow is from $V_b$ to $V_a$. 4. Combine R2 and R3 (series) to get the series branch resistance. $$R_{23}=R_2+R_3=5+20=25\,.$$ 5. Compute the branch currents using Ohm's law. Current through R1 (call it $I_{1}$): $$I_{1}=\frac{V_a-V_b}{R_1}=\frac{20-100}{10}=\frac{-80}{10}=\frac{-10\cdot 8}{10\cdot 1}=\frac{-\cancel{10}8}{\cancel{10}1}=-8\,.$$ This means $8$ flows from $V_b$ to $V_a$ (opposite the assumed direction). Current through the series R2--R3 branch (call it $I_{23}$; same through R2 and R3 because they are series): $$I_{23}=\frac{V_a-V_b}{R_{23}}=\frac{20-100}{25}=\frac{-80}{25}=\frac{-5\cdot 16}{5\cdot 5}=\frac{-\cancel{5}16}{\cancel{5}5}=-\frac{16}{5}=-3.2\,.$$ This means $3.2$ flows from $V_b$ to $V_a$ (opposite the assumed direction). 6. Voltages across each resistor (polarity left-to-right $V_a$ to $V_b$): Voltage across R1: $$V_{R1}=I_{1}R_1=(-8)\cdot 10=-80\,.$$ Voltage across R2: $$V_{R2}=I_{23}R_2=(-3.2)\cdot 5=-16\,.$$ Voltage across R3: $$V_{R3}=I_{23}R_3=(-3.2)\cdot 20=-64\,.$$ Check: $V_{R2}+V_{R3}=-16+(-64)=-80$, which equals $V_{R1}$, consistent with KVL between the two top nodes. 7. Currents through each element, summarized with directions (positive shown as left-to-right $V_a$ to $V_b$): $I_{R1}=I_{1}=-8$ (so 8 flows right-to-left from $V_b$ to $V_a$). $I_{R2}=I_{R3}=I_{23}=-3.2$ (so 3.2 flows right-to-left from $V_b$ to $V_a$ through R3 then R2). The currents through the ideal voltage sources are whatever values satisfy KCL at their connected nodes; for the left source V1 the source current is $I_{V1}=-(I_{R1}+I_{R2})= -(-8 + -3.2)=11.2$ flowing into the positive terminal of V1 by KCL, and for the right source V2 the source supplies $-(I_{V1})=-11.2$ to satisfy loop equilibrium. 8. Final numeric answers (magnitudes and polarities): Current through R1: 8 from right to left (numeric value $-8$ A if left-to-right sign convention used). Current through R2: 3.2 from right to left (numeric value $-3.2$ A left-to-right). Current through R3: 3.2 from right to left (numeric value $-3.2$ A left-to-right). Voltage across R1 (Va to Vb): -80 V. Voltage across R2 (Va to Vb): -16 V. Voltage across R3 (Va to Vb): -64 V.