1. **Find the equivalent resistance between terminals A and B for the two circuits:** (i) Circuit 1: - Resistors 12Ω and 4Ω are in series: $$R_{series1} = 12 + 4 = 16\ \Omega$$ - Resistors 8Ω and 12Ω are in series: $$R_{series2} = 8 + 12 = 20\ \Omega$$ - These two series groups (16Ω and 20Ω) are in parallel: $$\frac{1}{R_{eq}} = \frac{1}{16} + \frac{1}{20} = \frac{5}{80} + \frac{4}{80} = \frac{9}{80}$$ $$R_{eq} = \frac{80}{9} \approx 8.89\ \Omega$$ (ii) Circuit 2: - The 18Ω and 6Ω resistors in series: $$18 + 6 = 24\ \Omega$$ - Another 18Ω and 6Ω in series: $$18 + 6 = 24\ \Omega$$ - These two 24Ω resistors are in parallel: $$\frac{1}{R_{parallel}} = \frac{1}{24} + \frac{1}{24} = \frac{2}{24} = \frac{1}{12}$$ $$R_{parallel} = 12\ \Omega$$ - Add the 6Ω resistor in series: $$R_{eq} = 12 + 6 = 18\ \Omega$$ 2. a) **Ohm's Law:** - Ohm's law states that the voltage across a resistor is directly proportional to the current flowing through it, mathematically: $$V = IR$$ b) For the given circuit: (i) Resistance of entire circuit: - Resistors 6KΩ and 8KΩ in parallel: $$\frac{1}{R_1} = \frac{1}{6000} + \frac{1}{8000} = \frac{4}{24000} + \frac{3}{24000} = \frac{7}{24000}$$ $$R_1 = \frac{24000}{7} \approx 3428.57\ \Omega$$ - This is in series with 20KΩ: $$R_{total} = 3428.57 + 20000 = 23428.57\ \Omega$$ (ii) Current in each resistor: - Total current $$I = \frac{V}{R_{total}} = \frac{15}{23428.57} \approx 0.00064\ A$$ - Current through 6KΩ and 8KΩ (parallel branch): $$I_{6K} = I \times \frac{R_{other}}{R_{6K} + R_{other}} = 0.00064 \times \frac{8000}{6000 + 8000} = 0.00064 \times \frac{8000}{14000} \approx 0.00037\ A$$ $$I_{8K} = I - I_{6K} = 0.00064 - 0.00037 = 0.00027\ A$$ - Current through 20KΩ is same as total current: 0.00064 A (iii) Voltage drop across 10KΩ resistor: - Current through 10KΩ is same as current through that branch (assumed 0.00064 A) - Voltage drop: $$V = IR = 0.00064 \times 10000 = 6.4\ V$$ 3. Given current through 4Ω resistor is 3A. a) Current through other resistors: - Using Kirchhoff's laws and given currents: - Current through 3Ω resistor is same as 4Ω resistor branch: 3A - Current through 1Ω resistor is sum of currents through 3Ω and 4Ω resistors: $$I = 3 + 3 = 6A$$ b) Voltage of battery on the left: - Voltage drop across resistors: $$V_{total} = I_{1\Omega} \times 1 + I_{3\Omega} \times 3 + I_{4\Omega} \times 4 = 6 \times 1 + 3 \times 3 + 3 \times 4 = 6 + 9 + 12 = 27V$$ 4. For the RLC circuit: Given: $$R=500\ \Omega, L=2.4\ mH=2.4 \times 10^{-3} H, C=3.6\ \mu F=3.6 \times 10^{-6} F, V=240, f=60Hz$$ (a) Inductive reactance: $$X_L = 2\pi f L = 2 \times 3.1416 \times 60 \times 2.4 \times 10^{-3} = 0.9048\ \Omega$$ (b) Capacitive reactance: $$X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.1416 \times 60 \times 3.6 \times 10^{-6}} = 737.3\ \Omega$$ (c) Impedance: $$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{500^2 + (0.9048 - 737.3)^2} = \sqrt{250000 + (-736.4)^2} = \sqrt{250000 + 542282} = \sqrt{792282} \approx 890.1\ \Omega$$ (d) Phase angle: $$\theta = \tan^{-1} \left( \frac{X_L - X_C}{R} \right) = \tan^{-1} \left( \frac{0.9048 - 737.3}{500} \right) = \tan^{-1}(-1.472) = -55.3^\circ$$ (e) Supply current: $$I_s = \frac{V}{Z} = \frac{240}{890.1} = 0.2696\ A$$