2021 Series – Unit 1: Electrical Principles (Solved Past Papers & Model Answers)

2021 KNEC Examination Series: Electrical Principles Solved Questions

Subject: Electrical Principles (Module 1)
Series: July/November 2021 KNEC Examination
Status: Ready for Review & Approval

Question 1: DC Network Theorems & Kirchhoff’s Laws (20 Marks)

(a) State Kirchhoff’s Current Law (KCL) and Kirchhoff’s Voltage Law (KVL). [4 Marks]

• Kirchhoff’s Current Law (KCL): The algebraic sum of currents entering any electrical node or junction in a circuit is equal to zero (Sum of I_in = Sum of I_out). [2 Marks]
• Kirchhoff’s Voltage Law (KVL): The algebraic sum of all electromotive forces (EMFs) and potential differences around any closed loop in an electric circuit is equal to zero (Sum of V = Sum of I*R). [2 Marks]

(b) Two DC voltage sources E1 = 12 V (internal resistance r1 = 1 Ohm) and E2 = 10 V (internal resistance r2 = 2 Ohms) are connected in parallel across a load resistor RL = 5 Ohms. Using Thevenin’s Theorem, determine:
1. The Thevenin open-circuit voltage (V_th).
2. The Thevenin equivalent resistance (R_th).
3. The load current (I_L) and power dissipated in the load resistor. [16 Marks]

Model Solution & Derivations:

1. Calculate Thevenin Voltage (V_th):
   With load resistor RL removed from terminals A-B, circulating loop current between sources is:
   I_c = (E1 – E2) / (r1 + r2) = (12 – 10) / (1 + 2) = 2/3 A = 0.667 A.
   V_th = E1 – (I_c * r1) = 12 – (0.667 * 1) = 11.33 Volts. [5 Marks]
2. Calculate Thevenin Resistance (R_th):
   Deactivating independent voltage sources (short circuits):
   R_th = (r1 * r2) / (r1 + r2) = (1 * 2) / (1 + 2) = 0.67 Ohms. [4 Marks]
3. Calculate Load Current (I_L) & Power (P_L):
   I_L = V_th / (R_th + RL) = 11.33 / (0.67 + 5) = 11.33 / 5.67 = 2.00 Amperes. [4 Marks]
   P_L = (I_L)^2 * RL = (2.0)^2 * 5 = 20.0 Watts. [3 Marks]

Question 2: AC Series Resonance Calculations (20 Marks)

An R-L-C series circuit with R = 15 Ohms, L = 0.2 H, and C = 50 uF is connected to a 240 V, 50 Hz AC supply. Calculate:
1. Inductive reactance (X_L) and capacitive reactance (X_C).
2. Total circuit impedance (Z) and phase angle.
3. Resonant frequency (f_0). [20 Marks]

Model Solution:
– X_L = 2 * pi * f * L = 2 * pi * 50 * 0.2 = 62.83 Ohms.
– X_C = 1 / (2 * pi * f * C) = 1 / (2 * pi * 50 * 50*10^-6) = 63.66 Ohms.
– X_net = X_L – X_C = 62.83 – 63.66 = -0.83 Ohms (Capacitive).
– Z = sqrt(R^2 + (X_L – X_C)^2) = sqrt(15^2 + (-0.83)^2) = 15.02 Ohms.
– Resonant frequency f_0 = 1 / (2 * pi * sqrt(L * C)) = 1 / (2 * pi * sqrt(0.2 * 50*10^-6)) = 50.33 Hz.