1. **Problem Statement:** (a) For the first circuit, given the ammeter reading $I=1.25$ A, find the voltmeter reading $V$. (b) Find the emf $\varepsilon$ of the battery. 2. **Given Data:** Resistors: $R_1=25.0\ \Omega$, $R_2=15.0\ \Omega$, $R_3=45.0\ \Omega$, $R_4=15.0\ \Omega$, $R_5=10.0\ \Omega$, $R_6=35.0\ \Omega$. 3. **Step 1: Analyze the first circuit** - The ammeter reading $I=1.25$ A is the total current flowing through the circuit. - The voltmeter is connected across $R_1=25.0\ \Omega$ and $R_2=15.0\ \Omega$ in series. - Total resistance in that branch is $R_{12} = R_1 + R_2 = 25.0 + 15.0 = 40.0\ \Omega$. 4. **Step 2: Calculate voltage across $R_{12}$ (voltmeter reading):** $$V = I \times R_{12} = 1.25 \times 40.0 = 50.0\ \text{Volts}$$ 5. **Step 3: Calculate total resistance of the circuit to find emf $\varepsilon$** - The top branch resistors $R_3=45.0\ \Omega$ and $R_4=15.0\ \Omega$ are in series: $R_{34} = 45.0 + 15.0 = 60.0\ \Omega$. - The right branch resistors $R_5=10.0\ \Omega$ and $R_4=15.0\ \Omega$ are in series: $R_{54} = 10.0 + 15.0 = 25.0\ \Omega$. - The ammeter is in series with $R_{54}$. - The bottom resistor $R_6=35.0\ \Omega$ is in series with the parallel combination of $R_{34}$ and $R_{54}$. 6. **Step 4: Calculate equivalent resistance of parallel branches $R_{34}$ and $R_{54}$:** $$\frac{1}{R_p} = \frac{1}{R_{34}} + \frac{1}{R_{54}} = \frac{1}{60.0} + \frac{1}{25.0} = \frac{25 + 60}{60 \times 25} = \frac{85}{1500}$$ $$R_p = \frac{1500}{85} \approx 17.65\ \Omega$$ 7. **Step 5: Total resistance $R_{total}$:** $$R_{total} = R_{12} + R_p + R_6 = 40.0 + 17.65 + 35.0 = 92.65\ \Omega$$ 8. **Step 6: Calculate emf $\varepsilon$ using Ohm's law:** $$\varepsilon = I \times R_{total} = 1.25 \times 92.65 = 115.81\ \text{Volts}$$ --- 9. **Second circuit problem:** Given $R_1 = R_2 = 10.9\ \Omega$, $R_3 = 7.9\ \Omega$, and source voltage $V = 14.0$ V. 10. **Step 1: Calculate equivalent resistance of $R_1$ and $R_2$ in parallel:** $$\frac{1}{R_{12}} = \frac{1}{10.9} + \frac{1}{10.9} = \frac{2}{10.9}$$ $$R_{12} = \frac{10.9}{2} = 5.45\ \Omega$$ 11. **Step 2: Total resistance $R_{eq}$:** $$R_{eq} = R_{12} + R_3 = 5.45 + 7.9 = 13.35\ \Omega$$ 12. **Step 3: Total current $I$ from source:** $$I = \frac{V}{R_{eq}} = \frac{14.0}{13.35} \approx 1.048\ \text{A}$$ 13. **Step 4: Voltage across $R_3$:** $$V_3 = I \times R_3 = 1.048 \times 7.9 = 8.28\ \text{V}$$ 14. **Step 5: Voltage across parallel branch $R_{12}$:** $$V_{12} = V - V_3 = 14.0 - 8.28 = 5.72\ \text{V}$$ 15. **Step 6: Current through $R_1$ and $R_2$ (equal resistors):** $$I_1 = I_2 = \frac{V_{12}}{10.9} = \frac{5.72}{10.9} = 0.525\ \text{A}$$ --- 16. **Third problem: Light bulbs in series and parallel** Given resistances $R_a=400\ \Omega$, $R_b=800\ \Omega$, voltage $V=120$ V. 17. **(a) Current in series:** $$R_s = R_a + R_b = 400 + 800 = 1200\ \Omega$$ $$I_s = \frac{V}{R_s} = \frac{120}{1200} = 0.1\ \text{A}$$ 18. **(b) Power dissipated in each bulb in series:** $$P_a = I_s^2 \times R_a = 0.1^2 \times 400 = 4\ \text{W}$$ $$P_b = 0.1^2 \times 800 = 8\ \text{W}$$ 19. **(c) Total power in series:** $$P_{total} = P_a + P_b = 4 + 8 = 12\ \text{W}$$ 20. **(d) Current in parallel:** $$I_a = \frac{V}{R_a} = \frac{120}{400} = 0.3\ \text{A}$$ $$I_b = \frac{120}{800} = 0.15\ \text{A}$$ 21. **(e) Power dissipated in each bulb in parallel:** $$P_a = V^2 / R_a = \frac{120^2}{400} = 36\ \text{W}$$ $$P_b = \frac{120^2}{800} = 18\ \text{W}$$ 22. **(f) Total power in parallel:** $$P_{total} = 36 + 18 = 54\ \text{W}$$ 23. **(g) Brightness comparison:** - In series, bulb with higher resistance ($800\ \Omega$) dissipates more power. - In parallel, bulb with lower resistance ($400\ \Omega$) dissipates more power. 24. **(h) Greater total light output:** - Parallel connection dissipates more total power (54 W) than series (12 W), so parallel has greater light output.