Correct Option: **A** ### Step-by-step Solution #### 1. Calculate the total equivalent resistance of the circuit Since the resistors $R_1$ and $R_2$ are connected in series, their equivalent resistance ($R_{\text{eq}}$) is the sum of their individual resistances: $$R_{\text{eq}} = R_1 + R_2$$ Given $R_1 = 2\ \Omega$ and $R_2 = 4\ \Omega$: $$R_{\text{eq}} = 2\ \Omega + 4\ \Omega = 6\ \Omega$$ #### 2. Calculate the total current flowing through the circuit According to Ohm's Law, the total current ($I$) flowing through the circuit is given by the total voltage ($V$) divided by the total equivalent resistance ($R_{\text{eq}}$): $$I = \frac{V}{R_{\text{eq}}}$$ Given $V = 12\text{ V}$: $$I = \frac{12\text{ V}}{6\ \Omega} = 2\text{ A}$$ #### 3. Calculate the potential difference between point Q and point R Point Q is between $R_1$ and $R_2$, and point R is just after $R_2$ (before the negative terminal of the battery). Thus, the potential difference between Q and R ($V_{QR}$) is the potential difference across resistor $R_2$. Current flows from higher potential to lower potential, so $V_Q > V_R$: $$V_{QR} = V_Q - V_R = I \times R_2$$ Using the calculated current $I = 2\text{ A}$ and $R_2 = 4\ \Omega$: $$V_{QR} = 2\text{ A} \times 4\ \Omega = 8\text{ V}$$ #### 4. Calculate the work done by the electric field to move a positive charge from point Q to point R The work done by the electric field ($W$) to move a charge $q$ from an initial point with potential $V_i$ to a final point with potential $V_f$ is given by: $$W = -\Delta U = -(U_f - U_i) = -(qV_f - qV_i) = q(V_i - V_f)$$ Here, the initial point is Q ($V_i = V_Q$) and the final point is R ($V_f = V_R$). The charge is $q = 1\text{ C}$: $$W_{QR} = q(V_Q - V_R)$$ Using $q = 1\text{ C} $ and $V_Q - V_R = 8\text{ V}$: $$W_{QR} = 1\text{ C} \times 8\text{ V} = 8\text{ J}$$ Since the work done by the electric field is positive, it means the electric field naturally drives the charge from Q to R. --- ### Conclusion The total current flowing through the circuit is $2\text{ A}$, and the work done by the electric field to move a positive charge of $1\text{ C}$ from point Q to point R is $8\text{ J}$. **Evaluating the options:** * **A:** Matches our calculated values: Current $2\text{ A}$, Work done $8\text{ J}$. * **B:** Incorrect sign for work done. * **C:** Incorrect current and incorrect work done. * **D:** Incorrect current and incorrect work done. The final answer is **A**.