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 find the value of the resistor \( R_{1} \), we can use Kirchhoff's Voltage Law. The total voltage in the circuit can be expressed as the sum of the individual voltages across the resistors. From the information provided: 1. The total voltage \( E_{\mathrm{T}} = E_{1} + E_{2} + E_{3} \). 2. We can express \( E_{2} \) in terms of \( R_{1} \) and \( R_{2} \) using Ohm's law. Given: - \( E_{\mathrm{T}} = 233 \mathrm{~V} \) - \( E_{1} = 33 \mathrm{~V} \) - \( E_{3} = 24 \mathrm{~V} \) - \( R_{2} = 137 \Omega \) So, we can calculate \( E_{2} \): \[ E_{2} = E_{\mathrm{T}} - E_{1} - E_{3} = 233 \, \mathrm{V} - 33 \, \mathrm{V} - 24 \, \mathrm{V} = 176 \, \mathrm{V} \] Using Ohm's law, we know that: \[ E_{2} = I \cdot R_{2} \] The current \( I \) can also be expressed as: \[ I = \frac{E_{1}}{R_{1}} = \frac{E_{3}}{R_{3}} \] Thus, we can relate \( E_{2} \) to the resistances and voltage: \[ 176 \, \mathrm{V} = I \cdot 137 \, \Omega \] Now we can substitute \( I \) from \( E_{1} \): The equation for \( I \) from \( E_{1} \): \[ I = \frac{E_{1}}{R_{1}} = \frac{33 \, \mathrm{V}}{R_{1}} \] Substituting \( I \) in the equation of \( E_{2} \): \[ 176 = \frac{33}{R_{1}} \cdot 137 \] Now, solve for \( R_{1} \): \[ R_{1} = \frac{33 \cdot 137}{176} \] Calculating \( R_{1} \): \[ R_{1} = \frac{4521}{176} \approx 25.7 \, \Omega \] Rounding to one decimal place, the value of the resistor \( R_{1} \) is: \[ \boxed{25.7 \, \Omega} \]