Find the individual currents that flow through each of the following resistors: \( R_{1}=3 \Omega, R_{2}=5 \Omega, R_{3}=5 \Omega, R_{4}=10 \Omega \) \( \left(R_{2}\right. \) and \( R_{3} \) are parallel) \( E=24 \mathrm{~V} \)

To find the individual currents flowing through each resistor, we first need to determine the equivalent resistance of the circuit. Since \(R_2\) and \(R_3\) are in parallel, we can calculate their equivalent resistance \(R_{eq2-3}\) using the formula: \[ \frac{1}{R_{eq2-3}} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5} \implies R_{eq2-3} = \frac{5}{2} = 2.5 \, \Omega \] Now, we can find the total resistance \(R_{total}\) in the circuit, which includes \(R_1\) and the equivalent resistance of the two parallel resistors: \[ R_{total} = R_1 + R_{eq2-3} + R_4 = 3 \, \Omega + 2.5 \, \Omega + 10 \, \Omega = 15.5 \, \Omega \] Using Ohm’s law, we can now determine the total current \(I\) flowing through the circuit: \[ I = \frac{E}{R_{total}} = \frac{24 \, V}{15.5 \, \Omega} \approx 1.55 \, A \] Next, we need to find the voltage across the parallel resistors \(R_2\) and \(R_3\) using this current: \[ V_{parallel} = I \cdot R_{eq2-3} = 1.55 \, A \cdot 2.5 \, \Omega \approx 3.88 \, V \] Now, we can find the individual currents through the resistors \(R_2\) and \(R_3\): For \(R_2\): \[ I_2 = \frac{V_{parallel}}{R_2} = \frac{3.88 \, V}{5 \, \Omega} \approx 0.776 \, A \] For \(R_3\): \[ I_3 = \frac{V_{parallel}}{R_3} = \frac{3.88 \, V}{5 \, \Omega} \approx 0.776 \, A \] And finally, the current through \(R_1\) is the same as the total current since it’s in series: \[ I_1 = I \approx 1.55 \, A \] To summarize, the individual currents flowing through each resistor are: - \(I_1 \approx 1.55 \, A\) (through \(R_1\)) - \(I_2 \approx 0.776 \, A\) (through \(R_2\)) - \(I_3 \approx 0.776 \, A\) (through \(R_3\)) - Since \(R_4\) is also in series, \(I_4 = I \approx 1.55 \, A\) (though it's not specifically asked in the original question, it's worth noting as well).