1. **State the problem:** We need to find the value of resistor $R_1$ in a circuit where a voltage source $V_s = 97$ V supplies a total current $I_t = 522$ mA flowing through $R_1$. After $R_1$, the current splits into two branches with resistors $R_2 = 612$ Ohm and $R_3 = 1900$ Ohm (1.9 kOhm). 2. **Relevant formulas and rules:** - Ohm's Law: $$V = IR$$ relates voltage, current, and resistance. - The total current $I_t$ flows through $R_1$, so voltage across $R_1$ is $$V_{R_1} = I_t R_1$$. - After $R_1$, the current splits into $I_2$ and $I_3$ through $R_2$ and $R_3$ respectively. - The voltage across $R_2$ and $R_3$ is the same because they are in parallel: $$V_{R_2} = V_{R_3} = V_{out}$$. - Total current splits: $$I_t = I_2 + I_3$$. 3. **Calculate the equivalent resistance of $R_2$ and $R_3$ in parallel:** $$\frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{612} + \frac{1}{1900}$$ Calculate each term: $$\frac{1}{612} \approx 0.00163399, \quad \frac{1}{1900} \approx 0.00052632$$ Sum: $$0.00163399 + 0.00052632 = 0.00216031$$ So, $$R_{23} = \frac{1}{0.00216031} \approx 462.99 \text{ Ohm}$$ 4. **Calculate total resistance $R_{total}$:** Since $R_1$ is in series with the parallel combination $R_{23}$, $$R_{total} = R_1 + R_{23}$$ 5. **Use Ohm's law for the entire circuit:** $$V_s = I_t R_{total} = I_t (R_1 + R_{23})$$ Rearranged to solve for $R_1$: $$R_1 = \frac{V_s}{I_t} - R_{23}$$ 6. **Plug in values:** Convert $I_t$ to amperes: $$I_t = 522 \text{ mA} = 0.522 \text{ A}$$ Calculate: $$\frac{V_s}{I_t} = \frac{97}{0.522} \approx 185.82 \text{ Ohm}$$ Then, $$R_1 = 185.82 - 462.99 = -277.17 \text{ Ohm}$$ 7. **Interpretation:** A negative resistance is not physically possible here, indicating either the problem setup or given values may be inconsistent. However, mathematically, the calculated $R_1$ is: **Final answer:** $$R_1 = -277.17 \text{ Ohm}$$ This suggests a review of the circuit or values is needed, but this is the direct calculation based on given data.