1. **Problem Statement:** Find the currents through the 30.0-Ω resistor, 20.0-Ω resistor, and the 10.0-V battery in the circuit with two batteries (10.0 V and 5.00 V) and resistors 30.0 Ω and 20.0 Ω. 2. **Relevant Formulas and Rules:** - Use Kirchhoff's Voltage Law (KVL): The sum of voltage drops around any closed loop equals zero. - Ohm's Law: $V = IR$ where $V$ is voltage, $I$ current, and $R$ resistance. - Assume currents $I_1$ through 30.0 Ω and $I_2$ through 20.0 Ω. 3. **Set up equations:** Loop 1 (left loop with 10.0 V battery and 30.0 Ω resistor): $$10.0 - 30.0 I_1 - V_{junction} = 0$$ Loop 2 (right loop with 5.00 V battery and 20.0 Ω resistor): $$V_{junction} - 20.0 I_2 - 5.00 = 0$$ At junction, currents satisfy: $$I_1 = I_2$$ 4. **Solve for $V_{junction}$:** From loop 1: $$V_{junction} = 10.0 - 30.0 I_1$$ From loop 2: $$V_{junction} = 20.0 I_2 + 5.00$$ Set equal: $$10.0 - 30.0 I_1 = 20.0 I_2 + 5.00$$ Since $I_1 = I_2 = I$: $$10.0 - 30.0 I = 20.0 I + 5.00$$ 5. **Simplify and solve for $I$:** $$10.0 - 5.00 = 20.0 I + 30.0 I$$ $$5.00 = 50.0 I$$ $$I = \frac{5.00}{50.0} = 0.10$$ 6. **Answer for (a) and (b):** - Current through 30.0-Ω resistor: $I_1 = 0.10$ A - Current through 20.0-Ω resistor: $I_2 = 0.10$ A 7. **Current through 10.0-V battery (c):** The current through the 10.0-V battery is the same as $I_1$, which is $0.10$ A. --- 1. **Problem Statement:** Find the emfs $E_1$ and $E_2$ and the potential difference between points $b$ and $a$ in the circuit with given resistors and currents. 2. **Given:** - Currents: 1.00 A and 2.00 A - Resistors: 1.00 Ω, 4.00 Ω, 6.00 Ω, 1.00 Ω, 2.00 Ω - Voltage source: 20.0 V 3. **Use Kirchhoff's Laws and Ohm's Law:** - Calculate voltage drops using $V=IR$. - Use given currents and resistances to find $E_1$ and $E_2$. 4. **Calculate $E_1$:** Voltage drop across 4.00 Ω resistor with 1.00 A: $$V = 4.00 \times 1.00 = 4.00\text{ V}$$ Voltage drop across 1.00 Ω resistor with 2.00 A: $$V = 1.00 \times 2.00 = 2.00\text{ V}$$ Sum with 20.0 V source: $$E_1 = 20.0 + 4.00 + 2.00 = 26.0\text{ V}$$ 5. **Calculate $E_2$:** Voltage drop across 6.00 Ω resistor with 2.00 A: $$V = 6.00 \times 2.00 = 12.0\text{ V}$$ Voltage drop across 2.00 Ω resistor with 2.00 A: $$V = 2.00 \times 2.00 = 4.00\text{ V}$$ Sum: $$E_2 = 12.0 + 4.00 = 16.0\text{ V}$$ 6. **Potential difference $V_b - V_a$:** Sum voltage drops from $a$ to $b$: $$V_b - V_a = E_1 - (1.00 \times 1.00) - (4.00 \times 1.00) = 26.0 - 1.00 - 4.00 = 21.0\text{ V}$$ --- 1. **Problem Statement:** For a circuit with a 12-V cell, 10-kΩ resistor, and 20-μF capacitor, find: 1. Time constant 2. Charge at $t=0$ 3. Initial current 4. Capacitor charge at 0.08 min 5. Current at 0.08 min 2. **Formulas:** - Time constant: $$\tau = RC$$ - Charge on capacitor: $$q(t) = C V (1 - e^{-t/\tau})$$ - Current: $$I(t) = \frac{V}{R} e^{-t/\tau}$$ 3. **Calculate time constant:** $$R = 10,000\ \Omega,\ C = 20 \times 10^{-6} F$$ $$\tau = 10,000 \times 20 \times 10^{-6} = 0.2\text{ s}$$ 4. **Charge at $t=0$:** $$q(0) = C V (1 - e^{0}) = 0$$ 5. **Initial current:** $$I(0) = \frac{12}{10,000} e^{0} = 0.0012\text{ A} = 1.2\text{ mA}$$ 6. **Convert 0.08 min to seconds:** $$0.08 \times 60 = 4.8\text{ s}$$ 7. **Charge at $t=4.8$ s:** $$q(4.8) = 20 \times 10^{-6} \times 12 \times (1 - e^{-4.8/0.2})$$ Since $e^{-24}$ is approximately 0, $$q(4.8) \approx 20 \times 10^{-6} \times 12 = 2.4 \times 10^{-4} C$$ 8. **Current at $t=4.8$ s:** $$I(4.8) = \frac{12}{10,000} e^{-4.8/0.2} = 0.0012 \times e^{-24} \approx 0\text{ A}$$ **Final answers:** (a) $I_{30\Omega} = 0.10$ A (b) $I_{20\Omega} = 0.10$ A (c) Current through 10.0-V battery = 0.10 A $E_1 = 26.0$ V, $E_2 = 16.0$ V, $V_b - V_a = 21.0$ V Time constant $\tau = 0.2$ s Charge at $t=0$ is 0 C Initial current $I(0) = 1.2$ mA Charge at 0.08 min $= 2.4 \times 10^{-4}$ C Current at 0.08 min $\approx 0$ A