1. **State the problem:** We have a circuit with resistors $R_1=7\ \Omega$, $R_2=18\ \Omega$, $R_3=1\ \Omega$, $R_4=25\ \Omega$, and a voltage source $E=15\ V$. We want to find the current $I_E$ supplied by the voltage source and the power dissipated by resistor $R_2$, denoted $P_{R2}$. 2. **Analyze the circuit:** Resistors $R_3$ and $R_4$ are in parallel. Their equivalent resistance $R_{34}$ is given by: $$ R_{34} = \frac{R_3 R_4}{R_3 + R_4} = \frac{1 \times 25}{1 + 25} = \frac{25}{26} \ \Omega $$ 3. **Combine resistors:** $R_2$ is in series with $R_{34}$, so their combined resistance $R_{234}$ is: $$ R_{234} = R_2 + R_{34} = 18 + \frac{25}{26} = \frac{468}{26} + \frac{25}{26} = \frac{493}{26} \ \Omega $$ 4. **Total resistance in the circuit:** $R_1$ is in series with $R_{234}$, so total resistance $R_{total}$ is: $$ R_{total} = R_1 + R_{234} = 7 + \frac{493}{26} = \frac{182}{26} + \frac{493}{26} = \frac{675}{26} \ \Omega $$ 5. **Calculate current $I_E$ using Ohm's law:** $$ I_E = \frac{E}{R_{total}} = \frac{15}{\frac{675}{26}} = 15 \times \frac{26}{675} = \frac{390}{675} = \frac{\cancel{390}^{130}}{\cancel{675}^{225}} = \frac{130}{225} \approx 0.5778\ A $$ 6. **Calculate voltage across $R_2$:** Voltage across $R_{234}$ is: $$ U_{234} = I_E \times R_{234} = 0.5778 \times \frac{493}{26} = 0.5778 \times 18.96 \approx 10.95\ V $$ 7. **Calculate current through $R_2$:** Since $R_2$ is in series with $R_{34}$, current through $R_2$ is $I_E = 0.5778\ A$. 8. **Calculate power dissipated by $R_2$:** $$ P_{R2} = I_E^2 \times R_2 = (0.5778)^2 \times 18 = 0.3338 \times 18 = 6.01\ W $$ **Final answers:** $$ I_E \approx 0.578\ A, \quad P_{R2} \approx 6.01\ W $$