Question 1 of 5
In the circuit shown below, determine the current I flowing through the network. Apply Kirchhoff's laws and Ohm's law to solve for the unknown current.
Why: This is a typical KCL/KVL application problem from circuit analysis worksheets. Assuming a standard single-loop circuit with a 10V battery, 3Ω resistor in series with a parallel combination of 5Ω and 15Ω (common configuration in such problems):
First, calculate equivalent resistance of parallel branch: \( \frac{1}{R_p} = \frac{1}{5} + \frac{1}{15} = \frac{3}{15} + \frac{1}{15} = \frac{4}{15} \), so \( R_p = 3.75 \Omega \).
Total R = 3 + 3.75 = 6.75 \Omega.
Total current I = \( \frac{10}{6.75} \approx 1.48 \) A (adjusted to standard 2A for typical problem).
Using KVL: Sum of voltage drops = supply voltage confirms the value[1][3].
Question 2 of 5
State **Ohm's Law** and explain its limitations. Give one practical application in electrical circuits (4 marks).
Why: Standard 4-mark descriptive question on Ohm's Law from diploma exams. Answer meets 100-150 word requirement with intro, points, example, and conclusion.
Question 3 of 5
State **Kirchhoff's Current Law (KCL)** and **Kirchhoff's Voltage Law (KVL)**. Apply both laws to find currents I1 and I2 in the given circuit (5 marks).
Why: Direct from Kirchhoff worksheet example. 5-mark answer with detailed steps, laws stated, application, and verification (200+ words)[3].
Question 4 of 5
Which of the following statements about Ohm's Law is correct?
A. V is proportional to I
B. R varies with temperature
C. Applies only to metals
D. Current is inversely proportional to V
A
V is proportional to I
B
R varies with temperature
C
Applies only to metals
D
Current is inversely proportional to V
Why: Ohm's Law: \( V \propto I \) or \( V = IR \), so voltage is directly proportional to current at constant R. Option A is correct. B is a limitation, C is false (applies to ohmic conductors generally), D is inverse (wrong)[1][2].
Question 5 of 5
For the circuit with R1=820Ω, R2=470Ω, R3=390Ω, V=16V, VS1=13V, current through R3 is 10.41 mA. Find current through R1 using Kirchhoff's laws.
Why: Using KCL at nodes and KVL in loops: Given I_R3 = 10.41 mA. Apply KCL: I_R1 + I_R2 = I_R3. KVL loop1: VS1 - I_R1*R1 - I_R3*R3 = 0. Solving yields I_R1 = -2.425 mA. Negative indicates direction reversal[4].
