# Finding the Unknown Resistance \( R \) Given: -Capacitor charge:** \( Q = 350\, \mu\text{C} = 350 \times 10^{-6}\,\text{C} \) - **Capacitance:** \( C = 6\,\mu\text{F} = 6 \times 10^{-6}\,\text{F} \) - **Battery voltage:** \( V_\text{battery} = 300\,\text{V} \) ## 1. **Voltage Across the Capacitor** When fully charged, the voltage across the capacitor: \[ V_C = \frac{Q}{C} = \frac{350 \times 10^{-6}}{6 \times 10^{-6}} = 58.33\,\text{V} \] ## 2. **Circuit Analysis** When the capacitor is fully charged, **no current flows through it**, so the branch with the capacitor and 15Ω resistor is an open circuit. - The voltage across the capacitor also appears across the 15Ω resistor. ### **Path for analysis:** - Start at the bottom (ground), go through **25Ω**, then split: - One branch: **45Ω** → **15Ω + 6μF** (open) - Other branch: **30Ω** → **R** → join at the top. But with the capacitor fully charged, the branch with the capacitor and 15Ω is open, so all current flows **through the 45Ω, 30Ω, and R** only. #### Let's simplify step by step: ### **Voltage at the node between R and 15Ω** The potential difference across 15Ω is the same as across the capacitor: **58.33 V**. Let the top node (where 15Ω, R, 45Ω meet) be \( V_A \), and the bottom node (ground) be \( V_B \). - The voltage at \( V_A \) with respect to ground is **58.33 V**. ### **Total voltage drop** \[ V_\text{battery} = V_{25} + V_{45} + V_{R} + V_{30} \] But let's track the path that current flows: - **From battery (+):** up through 25Ω, splits at node: left (45Ω), right (30Ω + R), but left branch is open due to fully charged capacitor. So, **current only flows through:** - 25Ω → 30Ω → R → battery (-) But, **the point where R and 15Ω meet is at 58.33 V** above ground. Let’s call the current through 25Ω, 30Ω, and R as \( I \). ## 3. **Loop Equation** Let’s say the voltage at the node between R and 15Ω is \( V_A \). - \( V_A = 58.33\,\text{V} \) Starting from ground (battery negative), through 25Ω, through 30Ω, through R, up to node \( V_A \): \[ \xrightarrow{+300\,V} - I \cdot 25\Omega \xrightarrow{-I \cdot 30\Omega} - I \cdot R = V_A \] \[ 300\,\text{V} - I \cdot (25 + 30 + R) = 58.33\,\text{V} \] \[ I \cdot (55 + R) = 300 - 58.33 = 241.67 \] \[ I = \frac{241.67}{55 + R} \] ## 4. **Alternative Path (No Current)** The branch with 45Ω and (15Ω + capacitor) has no current, so ignore. ## 5. **Find R** But the key is to use the voltage at the node where R and 15Ω meet, which is 58.33 V above ground. - The voltage drop across the 30Ω resistor is \( V_{30} = I \cdot 30 \) - The voltage drop across the 25Ω resistor is \( V_{25} = I \cdot 25 \) - The voltage drop across R is \( V_{R} = I \cdot R \) Starting from the bottom (ground, V), after 25Ω, after 30Ω, after R, we reach \( V_A \): \[ + I \cdot 25 + I \cdot 30 + I \cdot R = V_A \] But from the battery's perspective: \[ 300\,\text{V} - I \cdot 25 - I \cdot 30 - I \cdot R = \] But, at \( V_A \), the voltage is 58.33 V. So, from ground to \( V_A \): \[ V_A = I \cdot 25 + I \cdot 30 + I \cdot R = I \cdot (55 + R) \] But from above, \( V_A = 58.33\,\text{V} \), and we also have \( I = \frac{241.67}{55 + R} \): \[ 58.33 = I \cdot (55 + R) \implies I = \frac{58.33}{55 + R} \] But previously, \( I = \frac{241.67}{55 + R} \). There seems to be a contradiction; let's clarify: From the **battery's total voltage**: \[ 300 = I \cdot (25 + 30 + R) + 58.33 \] \[ I \cdot (55 + R) = 241.67 \] \[ I = \frac{241.67}{55 + R} \] But, **the voltage at \( V_A \) is only from the drops across 25Ω and 30Ω**: \[ V_A = 300 - I \cdot 25 - I \cdot 30 = 300 - I \cdot 55 \] But \( V_A = 58.33 \): \[ 58.33 = 300 - I \cdot 55 \] \[ I \cdot 55 = 300 - 58.33 = 241.67 \] \[ I = \frac{241.67}{55} = 4.394 \,\text{A} \] Now, the voltage drop across R: \[ V_R = I \cdot R \] But from the previous equation: \[ 300 = I \cdot (25 + 30 + R) + 58.33 \] \[ 300 - 58.33 = I \cdot (55 + R) \] \[ 241.67 = I \cdot (55 + R) \] But we already found \( I = 4.394 \) A: \[ 241.67 = 4.394 \cdot (55 + R) \] \[ 55 + R = \frac{241.67}{4.394} = 55 \] \[ R = \,\Omega \] --- ## **Final Answer** ### **The resistance of the unknown resistor \( R \) is:** \[ \boxed{\,\Omega} \] --- ### **Summary of Reasoning** - The voltage at the node where R and 15Ω meet is fixed by the fully charged capacitor: 58.33 V. - The total voltage from the battery minus this node voltage gives the voltage drop across 25Ω, 30Ω, and R. - Calculations show that R must be Ω for the voltage distribution to match the given charge on the capacitor. --- **If you have any questions or would like to discuss alternative scenarios, let me know!**