LESSON 04 – PARALLEL CIRCUITS

WORK EXAMPLE SOLUTIONS

EXAMPLE #1 SOLUTION

What is the total resistance in the following simple parallel circuits?

a)		Solution (keep high precision till the end) RT = (1/R1 + 1/R2)^-1 RT = (1/200 + 1/100)^-1 RT = (0.005 + 0.01)^-1 RT = (0.015)^-1 RT = 66.7 ohms
b)		Solution (keep high precision till the end) RT = (1/R1 + 1/R2 + 1/R3)^-1 RT = (1/250 + 1/400 + 1/120)^-1 RT = (0.004 + 0.0025 + 0.00833333)^-1 RT = (0.0148333)^-1 RT = 67.4 ohms

EXAMPLE #2 SOLUTION

Consider the following circuit (same as 1a).

STEP 1 – We start with the table.

VT = 12V	IT	RT
V1	I1	R1 = 200Ω
V2	I2	R2 = 100Ω

STEP 2 – Since this is the same problem as 1a, we know that RT = 66.7Ω.

VT = 12V	IT	RT = 66.7Ω
V1	I1	R1 = 200Ω
V2	I2	R2 = 100Ω

STEP 3 – We can now solve for I_T.

V_T = I_T x R_T

12V = I_T x 66.67W

12V / 66.7W = I_T

0.18A = I_T

VT = 12V	IT = 0.18A	RT = 66.7Ω
V1	I1	R1 = 200Ω
V2	I2	R2 = 100Ω

STEP 4 – We now need to use our knowledge of parallel circuits to figure out another unknown.

We know that the voltage drop in each branch is the same as V_T.

So, V₁ = V₂ = V_T = 12V

VT = 12V	IT = 0.18A	RT = 66.7Ω
V1 = 12V	I1	R1 = 200Ω
V2 = 12V	I2	R2 = 100Ω

STEP 5 – We can now solve for I₁.

V₁ = I₁ x R₁

12V = I₁ x 200W

12V / 200W = I₁

0.06A = I₁

STEP 6 – We can now solve for I₂.

V₂ = I₂ x R₂

12V = I₂ x 100W

12V / 100W = I₂

0.12A = I₂

VT = 12V	IT = 0.18A	RT = 66.7Ω
V1 = 12V	I1 = 0.06A	R1 = 200Ω
V2 = 12V	I2 = 0.12A	R2 = 100Ω

We are done. A quick check shows that I₁ + I₂ is in fact equal to I_T so that suggests we did our work correctly.