Determine the value of the resistor, \( R_{1} \), given: \( E_{\mathrm{T}}=233 \mathrm{~V}, E_{1}=33 \mathrm{~V}, E_{3}=24 \mathrm{~V} \), and \( R_{2}=137 \Omega \) (Round the FINAL answer to one decimal place.)

To determine the value of the resistor \( R_{1} \), we can use Kirchhoff's voltage law, which states that the sum of the voltages around a closed loop must equal zero. Given the values: - \( E_{\mathrm{T}} = 233 \, \text{V} \) - \( E_{1} = 33 \, \text{V} \) - \( E_{3} = 24 \, \text{V} \) - \( R_{2} = 137 \, \Omega \) We can set up the equation based on the voltage drops across the resistors and the voltage sources: \[ E_{\mathrm{T}} = E_{1} + E_{2} + E_{3} \] Where \( E_{2} \) is the voltage drop across \( R_{1} \) and \( R_{2} \). The voltage drop across \( R_{2} \) can be calculated using Ohm's law: \[ E_{2} = I \cdot R_{2} \] To find the current \( I \), we can express it in terms of the voltages: \[ I = \frac{E_{1}}{R_{1}} = \frac{E_{3}}{R_{2}} \] Now, substituting \( I \) into the equation for \( E_{2} \): \[ E_{2} = \frac{E_{1}}{R_{1}} \cdot R_{2} \] Now substituting \( E_{2} \) back into the Kirchhoff's equation: \[ E_{\mathrm{T}} = E_{1} + \frac{E_{1}}{R_{1}} \cdot R_{2} + E_{3} \] Rearranging gives: \[ E_{\mathrm{T}} - E_{1} - E_{3} = \frac{E_{1}}{R_{1}} \cdot R_{2} \] Now we can solve for \( R_{1} \): \[ R_{1} = \frac{E_{1} \cdot R_{2}}{E_{\mathrm{T}} - E_{1} - E_{3}} \] Now substituting the known values: \[ R_{1} = \frac{33 \cdot 137}{233 - 33 - 24} \] Calculating the denominator: \[ 233 - 33 - 24 = 176 \] Now substituting back into the equation: \[ R_{1} = \frac{33 \cdot 137}{176} \] Calculating \( R_{1} \): \[ R_{1} = \frac{4521}{176} \approx 25.7 \] Rounding to one decimal place, we find: \[ R_{1} \approx 25.7 \, \Omega \] Thus, the value of the resistor \( R_{1} \) is approximately \( 25.7 \, \Omega \).