1. Problem 26.6 (a) and (b): Given a circuit with an ammeter reading 1.25 A, find the voltmeter reading and the emf $\mathcal{E}$ of the battery. 2. Use Ohm's Law and series-parallel resistor rules. The voltmeter is across the 45.0 $\Omega$ resistor, so voltage $V = IR$. 3. Calculate voltage across 45.0 $\Omega$ resistor: $$V = 1.25 \times 45.0 = 56.25\text{ V}$$ 4. To find emf $\mathcal{E}$, sum voltage drops in the loop. Total resistance $R_{total} = 45.0 + 25.0 + 15.0 + 15.0 + 10.0 + 35.0 = 145.0\ \Omega$. 5. Calculate emf: $$\mathcal{E} = I \times R_{total} = 1.25 \times 145.0 = 181.25\text{ V}$$ --- 6. Problem 2: For the circuit with $R_1 = R_2 = 10.9\ \Omega$, $R_3 = 7.9\ \Omega$, and battery voltage 14.0 V, find equivalent resistance, voltmeter, and ammeter readings. 7. $R_1$ and $R_2$ are in series: $$R_{12} = R_1 + R_2 = 10.9 + 10.9 = 21.8\ \Omega$$ 8. $R_{12}$ is in parallel with $R_3$: $$\frac{1}{R_{eq}} = \frac{1}{R_{12}} + \frac{1}{R_3} = \frac{1}{21.8} + \frac{1}{7.9}$$ 9. Calculate $R_{eq}$: $$\frac{1}{R_{eq}} = 0.0459 + 0.1266 = 0.1725 \Rightarrow R_{eq} = \frac{1}{0.1725} = 5.8\ \Omega$$ 10. Total current from battery: $$I = \frac{V}{R_{eq}} = \frac{14.0}{5.8} = 2.41\text{ A}$$ 11. Voltage across $R_3$: $$V_3 = I \times R_3 = 2.41 \times 7.9 = 19.04\text{ V}$$ (Check circuit configuration; if parallel, voltage across $R_3$ equals voltage across $R_{12}$.) 12. Current through $R_3$: $$I_3 = \frac{V}{R_3} = \frac{14.0}{7.9} = 1.77\text{ A}$$ 13. Current through $R_{12}$: $$I_{12} = I - I_3 = 2.41 - 1.77 = 0.64\text{ A}$$ 14. Voltage across $R_{12}$: $$V_{12} = I_{12} \times R_{12} = 0.64 \times 21.8 = 13.95\text{ V}$$ 15. Voltmeter reading is voltage across $R_3$ or $R_{12}$, approximately 14.0 V. --- 16. Problem 26.21: Two bulbs with resistances 400 $\Omega$ and 800 $\Omega$ connected to 120 V. (a) Series current: $$I = \frac{V}{R_1 + R_2} = \frac{120}{400 + 800} = \frac{120}{1200} = 0.1\text{ A}$$ (b) Power in each bulb: $$P_1 = I^2 R_1 = 0.1^2 \times 400 = 4\text{ W}, \quad P_2 = 0.1^2 \times 800 = 8\text{ W}$$ (c) Total power: $$P_{total} = P_1 + P_2 = 12\text{ W}$$ (d) Parallel currents: $$I_1 = \frac{V}{R_1} = \frac{120}{400} = 0.3\text{ A}, \quad I_2 = \frac{120}{800} = 0.15\text{ A}$$ (e) Power in each bulb: $$P_1 = V I_1 = 120 \times 0.3 = 36\text{ W}, \quad P_2 = 120 \times 0.15 = 18\text{ W}$$ (f) Total power: $$P_{total} = 36 + 18 = 54\text{ W}$$ (g) Brightest bulb: In series, 800 $\Omega$ bulb dissipates more power; in parallel, 400 $\Omega$ bulb dissipates more power. (h) Greater total light output is in parallel connection (54 W vs 12 W).