1. **Problem Statement:** Calculate the indicated currents and voltage for the given circuit (Fig. 7) which includes resistors $R_1=4k\Omega$, $R_2=8k\Omega$, $R_3=12k\Omega$, $R_4=24k\Omega$, $R_5=12k\Omega$, $R_6=12k\Omega$, $R_7=9k\Omega$, $R_8=3k\Omega$, $R_9=6k\Omega$, and a battery $E=72V$. Currents $I_s$ are indicated between points $a$ and the positive terminal of the battery (upward) and from $b$ rightward. $V_7$ is the voltage across $R_7$. 2. **Analyze Left Loop (Resistors $R_1$ to $R_5$ and battery):** - Combine series and parallel resistors to find the equivalent resistance. - $R_2$ and $R_3$ are in series: $$R_{23} = R_2 + R_3 = 8k + 12k = 20k\Omega$$ - $R_1$ in parallel with $R_{23}$: $$\frac{1}{R_{123}} = \frac{1}{R_1} + \frac{1}{R_{23}} = \frac{1}{4k} + \frac{1}{20k} = \frac{5 + 1}{20k} = \frac{6}{20k} \Rightarrow R_{123} = \frac{20k}{6} = 3.33k\Omega$$ - Then $R_4$ and $R_5$ are in series: $$R_{45} = R_4 + R_5 = 24k + 12k = 36k\Omega$$ - Total resistance of left loop, $R_{left} = R_{123} + R_{45} = 3.33k + 36k = 39.33k\Omega$ 3. **Calculate current $I_s$ through the battery (left loop):** $$I_s = \frac{E}{R_{left}} = \frac{72V}{39330\Omega} \approx 0.00183 A = 1.83 mA$$ 4. **Analyze Right Triangular Loop (Resistors $R_6$ to $R_9$):** - Resistors $R_8$ and $R_9$ are in series: $$R_{89} = R_8 + R_9 = 3k + 6k = 9k\Omega$$ - $R_7$ and $R_{89}$ are in parallel: $$\frac{1}{R_{789}} = \frac{1}{R_7} + \frac{1}{R_{89}} = \frac{1}{9k} + \frac{1}{9k} = \frac{2}{9k} \Rightarrow R_{789} = \frac{9k}{2} = 4.5k\Omega$$ - Total right loop resistance considering $R_6$ in series: $$R_{right} = R_6 + R_{789} = 12k + 4.5k = 16.5k\Omega$$ 5. **Calculate the current $I_b$ (rightward current from b):** - Assuming voltage at $b$ is ground (0V) and connected to the battery: current depends on the potential difference but lacking full connection details, let's assume the battery drives the same voltage $E$ across this part: $$I_b = \frac{E}{R_{right}} = \frac{72V}{16500\Omega} \approx 0.00436 A = 4.36 mA$$ 6. **Calculate voltage $V_7$ across $R_7$:** - Current through $R_7$ is same as current through $R_{789}$ branch. Since $R_7$ and $R_{89}$ are in parallel, current splits equally because their resistances are equal. So: - Current through $R_7$: $$I_7 = \frac{I_b}{2} = \frac{4.36mA}{2} = 2.18 mA$$ - Voltage across $R_7$: $$V_7 = I_7 \times R_7 = 2.18mA \times 9k\Omega = 0.00218 A \times 9000 \Omega = 19.62 V$$ **Final answers:** - $I_s = 1.83 mA$ (upward between $a$ and battery positive) - $I_b = 4.36 mA$ (rightward from $b$) - $V_7 = 19.62 V$ (across $R_7$ from minus to plus) These steps carefully combine resistors and apply Ohm's law and parallel/series current rules to find indicated values.