LESSON 05 – COMBINATION CIRCUITS

WORK EXAMPLE SOLUTIONS

EXAMPLE #1 SOLUTION

We have to answer questions related to this circuit:

a) The value of I₄ is the same as the value of I_T.

Answer: True

Explanation:
Since all electrons must go through R₄, we know that the current at R₄ is the same as the current leaving the battery. So, I₄ = I_T.

b) The value of I3 is the same as the value of I_T.

Answer: False

Explanation:
Some of the current leaving the battery will go into the top branch of the parallel section. Therefore, the current at R₃ will be less than the current leaving the battery. So, I₃ cannot be equal to I_T.

c) The value of I₁ is the same as the value of I₃.

Answer: False

Explanation:
The resistor R₁ and R₃ are on different paths. The only way they could have the same amount of current going through them is if the resistance of each branch is the same (splitting the overall current equally among both branches). Since the resistance in branch 12 is 350 ohms which is different than the resistance in the branch with R₃ which is 200 ohms, we know that the current cannot be the same in both branches. So, I₁ is not equal to I₃.

d) The value of I₁ is the same as the value of I₂.

Answer: True

Explanation:
The resistors R1 and R2 are on the same branch. So all current that passes through R1 will pass through R2. So I1 equals I2.

e) The resistance of branch 12 is 350 ohms.

Answer: True

Explanation:
The branch contains two resistors that are in series. So, to calculate the resistance of the branch, we simply add up the individual resistances. So, RB12 is equal to 100 + 250 = 350 ohms.

f) The resistance of the parallel section is more than 200 ohms.

Answer: False

Explanation:
We can solve this by solving the resistance of the parallel section. However, we can also do this by observation. Let’s imagine that we disconnected the branch B12. Then, the resistance of that parallel section would 200 ohms. Now, by adding the branch B12 back, since we are giving a new path for current to travel, we are reducing the resistance. So, the resistance of the parallel circuit has to be less than 200 ohms.

EXAMPLE #2 SOLUTION (SINGLE TABLE APPROACH)

We have to solve for all unknowns in the following circuit:

STEP 1 – We start with the table.

VT = 9V	IT	RT
V1	I1	R1 = 100Ω
V2	I2	R2 = 250Ω
V3	I3	R3 = 200Ω

STEP 2 – Because this is a combination circuit, we can no longer use the series or parallel rules that helped us solve the column unknowns. So, to solve for RT, we have to analyze the situation.

The resistance of the top branch is 250 + 100 = 350Ω because the two resistors are in series.

If we were to replace R₁ and R₂ by a 350Ω resistor, we would have a parallel circuit and we could easily solve for R_T.

R_T = (1/350 + 1/200)^-1

R_T = (0.00285714 + 0.005)^-1

R_T = (0.00785714)^-1

R_T = 127.27 W

VT = 9V	IT	RT = 127.27 W
V1	I1	R1 = 100Ω
V2	I2	R2 = 250Ω
V3	I3	R3 = 200Ω

STEP 3 – We can solve for I_T.

V_T = I_T x R_T

9V = I_T x 127.27W

9V/127.27W = I_T

0.0707A = I_T

VT = 9V	IT = 0.0707A	RT = 127.27 W
V1	I1	R1 = 100Ω
V2	I2	R2 = 250Ω
V3	I3	R3 = 200Ω

STEP 4 – We again need to use analysis to figure out how to proceed.

Electrons that pass through R₃ only go through that load. So they have to lose 9V. So V₃ is equal to 9V.

VT = 9V	IT = 0.0707A	RT = 127.27 W
V1	I1	R1 = 100Ω
V2	I2	R2 = 250Ω
V3 = 9V	I3	R3 = 200Ω

STEP 5 – We can solve for I₃.

V₃ = I₃ x R₃

9V = I₃ x 200W

9V/200W = I₃

0.045A = I₃

VT = 9V	IT = 0.0707A	RT = 127.27 W
V1	I1	R1 = 100Ω
V2	I2	R2 = 250Ω
V3 = 9V	I3 = 0.045A	R3 = 200Ω

STEP 6 – It’s time to analyze the circuit again.

We see that all electrons leaving the source either go through the branch with R1 and R2, or go through the branch with R3.

Let I12 be the current in the branch with R1 and R2.

I_T = I₁₂ + I₃

I₁₂ = I_T – I₃

I₁₂ = 0.0707A - 0.045A

I₁₂ = 0.0257A

Now that we know the current going through the branch with R1 and R2, we have the current going through R1 and R2 as well.

I₁ = I₂ = I₁₂ = 0.0257A

VT = 9V	IT = 0.0707A	RT = 127.27 W
V1	I1 = 0.0257A	R1 = 100Ω
V2	I2 = 0.0257A	R2 = 250Ω
V3 = 9V	I3 = 0.045A	R3 = 200Ω

STEP 7 – We can now solve for V₁ and V₂.

V₁ = I₁ x R₁

V₁ = 0.0257A x 100W

V₁ = 2.57 V

V₂ = V_T – V₁

V₂ = 9V – 2.57V

V₂ = 6.43V

VT = 9V	IT = 0.0707A	RT = 127.27 W
V1 = 2.57V	I1 = 0.0257A	R1 = 100Ω
V2 = 6.43V	I2 = 0.0257A	R2 = 250Ω
V3 = 9V	I3 = 0.045A	R3 = 200Ω

And that’s it!

EXAMPLE #3

We are asked to find the voltage drop at R1 and R2 without first figuring out the current in that branch. So we have to use ratios.

STEP 1 - The total resistance in the branch is 350 ohms.

STEP 2 – We can calculate the percentage that R₁ contributes to the branch’s total resistance.

R₁ represents 100/350 = 28.57% of the resistance in the branch.

STEP 3 – We know that an electron going through the branch with R₁ and R₂ will lose 9V. So it will lose 28.57% of that 9V at R₁.

V₁ = 9V x 28.57% = 2.57V

STEP 4 – The voltage drop over R₂ can now be calculated.

V₂ = 9V – 2.57V = 6.43V.