Basic Circuit Problem (2)

1) Find V1, V2 and V3 (in V) on the circuit shown below (Figure#1)

Figure#1

Circuit Analysis: Step-by-Step

Given:
V_a = 10 V, R_a = 5 kΩ, R_b = 2 kΩ, R_c = 2 kΩ
Note: R_b and R_c are in parallel.

Step 1: Equivalent Resistance of Parallel Branch

R_bc = R_b || R_c = (R_b × R_c) / (R_b + R_c)

R_bc = (2 × 2) / (2 + 2) = 1 kΩ

Step 2: Total Current in the Circuit

R_total = R_a + R_bc = 5 + 1 = 6 kΩ

I = V_a / R_total = 10 / 6 ≈ 1.667 mA

Step 3: Voltage across R_a (V₁)

V₁ = I × R_a = 1.667 × 5 ≈ 8.33 V

Step 4: Voltage across R_b and R_c (V₂ and V₃)

V_bc = V_a - V₁ = 10 - 8.33 ≈ 1.67 V

Since they are in parallel: V₂ = V₃ = 1.67 V

Final Results

Parameter	Calculated Value
Voltage V₁ (across R_a)	8.33 V
Voltage V₂ (across R_b)	1.67 V
Voltage V₃ (across R_c)	1.67 V

2) Find V1, V2 (in V) and I1, I2 (in mA) on the circuit shown below (Figure#2)

Figure#2

Circuit Solution

Note on Circuit Path: The branch with I₁ is a short circuit to ground. This means all current flows through that wire, bypassing R_c and R_d entirely.

1. Total Resistance (R_total)

Only R_a and R_b are active in the loop:

R_total = R_a + R_b = 3 kΩ + 2 kΩ = 5 kΩ

2. Total Current (I₁)

I₁ = V_a / R_total = 10V / 5 kΩ = 2 mA

3. Branch Current (I₂)

Since the current is diverted by the short circuit: I₂ = 0 mA

4. Voltages (V₁ and V₂)

V₁ = I₁ × R_a = 2mA × 3 kΩ = 6 V

V₂ = I₂ × R_d = 0mA × 5 kΩ = 0 V

Variable	Result
V₁	6 V
V₂	0 V
I₁	2 mA
I₂	0 mA

[Back]