1. **State the problem:** Given the node voltages $V_A=7$ V and $V_B=2$ V, and the voltage drop across resistor $R$ as $V_R=2$ V, we need to find the resistance value of $R$. 2. **Analyze the circuit portion around $R$:** The voltage drop $V_R$ is across $R$ with polarity given as $- (V_R) + (V_R)$. The voltage drop across $R$ is given by $V_A - V_B$ if $R$ is between nodes $A$ and $B$. 3. **Check given voltage drop across $R$:** Given $V_R=2$ V but $V_A - V_B = 7 - 2 = 5$ V. This suggests $R$ is not just the resistor between nodes $A$ and $B$ in the top branch but includes a different position. 4. **Identify $R$ and $V_R$ position:** The diagram shows $R$ connected possibly between the lower branches with its voltage drop $V_R=2$ V. Considering $V_R$ is across $R$ with polarity (-) to (+), so current $I$ flows from higher to lower potential. 5. **Calculate current through $R$: ** Using Ohm's Law $V=IR$, if we knew current $I$ through $R$, resistance would be $R = \frac{V_R}{I}$. 6. **Analyze node voltage to find current $I_2$: ** Given the resistor values and node voltages, currents can be calculated using Kirchhoff's Voltage Law (KVL) or Ohm's Law on branches. 7. **Calculate voltage difference across known 4Ω resistor after $V_B$: ** $$V_{4\Omega} = V_B - V_{ground} = 2 - 0 = 2 \text{ V}$$ 8. **Calculate current through 4Ω resistor $I_2$: ** $$I_2 = \frac{V_{4\Omega}}{4} = \frac{2}{4} = 0.5 \text{ A}$$ 9. **Assuming same current passes through $R$: ** Given $V_R = I_R \times R$, and $I_R = 0.5$ A (same as $I_2$), we get $$R = \frac{V_R}{I_R} = \frac{2}{0.5} = 4 \ \Omega$$ **Final answer:** The resistance value $R$ is **4 Ω**.