1. Problem: Calculate the equivalent resistance $R_{eq}$ for the circuit in Fig. 2.93, where all resistors are 5 $\Omega$ each and connected as shown. 2. Since the four 5 $\Omega$ resistors are connected in parallel, we use the formula for parallel resistors: $$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}$$ 3. Substitute each $R_i = 5$ $\Omega$: $$\frac{1}{R_{eq}} = \frac{1}{5} + \frac{1}{5} + \frac{1}{5} + \frac{1}{5} = \frac{4}{5}$$ 4. Solve for $R_{eq}$: $$R_{eq} = \frac{5}{4} = 1.25 \ \Omega$$ --- 1. Problem: For the circuit in Fig. 2.90 where $i_o = 3$ A, calculate $i_x$ and the total power absorbed by the circuit. 2. Given resistors: 10 $\Omega$, 8 $\Omega$, 4 $\Omega$, 2 $\Omega$, 16 $\Omega$. Currents $i_o = 3$ A and $i_x$ unknown. 3. Assuming the current division rule and resistor positions, $i_x$ flows through the 10 $\Omega$ resistor, and $i_o$ is total output current. 4. By Kirchhoff's current law and circuit analysis, find $i_x$ proportional to resistance ratios (specific details rely on circuit layout). Since full circuit connection details or voltages are not given, we cannot precisely compute $i_x$ or power. --- 1. Problem: Find the equivalent resistance $R_{eq}$ for the circuit in Fig. 2.94 with resistors 25 $\Omega$, 180 $\Omega$, 60 $\Omega$, and 60 $\Omega$ in series. 2. For series resistors, equivalant resistance is the sum: $$R_{eq} = R_1 + R_2 + R_3 + R_4$$ 3. Substitute values: $$R_{eq} = 25 + 180 + 60 + 60 = 325 \ \Omega$$