Resistors in Parallel and in Series Circuits Problems and Solutions

 Problem #1

Given the following series circuit, find: (a) the total resistance, (b) the total current, (c) the current through each resistor, (d) the voltage across each resistor, (e) the total power, (f) the power dissipated by each resistor!

Answer;
Known:
V = 24 V
R₁ = 3 Ω
R₂ = 5 Ω
R₃ = 4 Ω

(a) Total resistance:

R_T = R₁ + R₂ + R₃

R_T = 3 Ω + 5 Ω + 4 Ω = 12 Ω

(b) the total current

i = V/R_T = 24 V/12 Ω =  2 A

(c) the current through each resistor,

You know that the total current is 2 A. In a series circuit, i₁ = i₂ = i₃, so the current through each resistor is 2 A.

(d) the voltage across each resistor.

Once you know two of the four variables (V, I, P, R), you can find the other two variables. In this case, you know I and R and can find the voltage using Ohm’s law for each resistor.

V₁ = i₁R₁ = 2 A x 3 Ω = 6 V
V₂ = i₂R₂ = 2 A x 5 Ω = 10 V
V₃ = i₃R₃ = 2 A x 4 Ω = 8 V

(e) the total power is

P_T = V_Ti = 24 V x 2 A = 48 Watt

(f) the power dissipated by each resistor is

P₁ = V₁i₁ = 6 V x 2 A = 12 Watt
P₂ = V₂i₂ = 10 V x 2 A = 20 Watt
P₃ = V₃i₃ = 8 V x 2 A = 16 Watt

Problem #2
Given the following parallel circuit, find: (a) the current through each resistor, (b) the total current (c) the total power (d) the power in each resistor, (e) the total resistance.

Answer;
Known:
V = 24 V
R₁ = 8 Ω
R₂ = 6 Ω
R₃ = 12 Ω

(a) Total resistance:

1/R_T = 1/R₁ + 1/R₂ + 1/R₃

1/R_T = 1/8 Ω + 1/6 Ω + 1/12 Ω = 18/48 Ω

R_T = 2.67 Ω

(b) the total current

i = V/R_T = 24 V/(2.67 Ω) =  9.0 A

(c) the voltage across each resistor.

This is a parallel circuit. The current follows different paths to each resistor. In a parallel circuit, the voltage drops across each resistor are equal. In this case, the voltage of each resistor equals 24 V.
V₁ = V₂ = V₃ = V

You can put this information in the diagram immediately. Once you know two of the four variables (V, I, P, R), you can find the other two variables. In this case, you know V and R. You can find the current using Ohm’s law for each resistor.

(d) the current through each resistor,

i₁ = V/R₁ = 24 V/8 Ω = 3.0 A
i₂ = V/R₂ = 24 V/6 Ω = 4.0 A
i₃ = V/R₃ = 24 V/12 Ω = 2.0 A

(e) the total power is

P_T = V_Ti = 24 V x 9 A = 216 Watt

(f) the power dissipated by each resistor is

P₁ = V₁i₁ = 24 V x 3.0 A = 72 Watt
P₂ = V₂i₂ = 24 V x 4.0 A = 96 Watt
P₃ = V₃i₃ = 24 V x 2 A = 48 Watt

Problem #3
Shown below is a series/parallel circuit. (a) Calculate the total resistance of the series/parallel circuit shown below. (b) Calculate the current through and voltage across each each resistor.

Answer;
Known:
V = 24 V
R₁ = 2 Ω
R₂ = 10 Ω
R₃ = 15 Ω

(a) the total resistance of the series/parallel circuit shown below.

• R₂ and R₃ arranged in parallel,

R_p = R₂R₃/(R₂ + R₃) = (10 Ω)(15 Ω)/(10 Ω + 15 Ω) = 6 Ω

• R₁ and R_p arranged in series, then

R_T = R₁ + R_p = 2 Ω + 6 Ω = 8 Ω

(b) the current through each resistor

the total current is, i_T = V/R_T = 24 V/8 Ω = 3 A

i_T pass R₁, then i₁ = i_T = 3 A

the voltage on R₁ is V₁ = i₁R₁ = (3 A)(2 Ω) = 6 V

so the voltages on R₂ and R₃ are arranged in parallel the same magnitude, i.e.

V₂ = V₃ = V ─ V₁ = 24 V ─ 6 V = 18 V

then the magnitude of current i₂ and i₃ is

i₂ = V₂/R₂ = 18 V/10 Ω = 1.8 A
i₃ = V₃/R₃ = 18 V/15 V = 1.2 A

Problem #4
Shown below is a series/parallel circuit. (a) Calculate the total resistance of the series/parallel circuit shown below. (b) Calculate the current through and voltage across each each resistor.

Answer;
Known:
V = 24 V
R₁ = 2.4 Ω
R₂ = 8 Ω
R₃ = 12 Ω
R₄ = 6 Ω
R₅ = 24 Ω

(a) the total resistance of the series/parallel circuit shown below

• R₂ is in parallel with R₃

R₂₃ = R₂R₃/(R₂ + R₃) = (8 Ω)(12 Ω)/(8 Ω + 12 Ω) = 4.8 Ω

• R₄ is in parallel with R₅,

R₄₅ = R₄R₅/(R₄ + R₅) = (6 Ω)(24 Ω)/(6 Ω + 24 Ω) = 7.2 Ω

• R₁, R₂₃ and R₄₅ the series combination, then

R_T = R₁ + R₂₃ + R₄₅ = 4 Ω + 4.8 Ω + 7.2 Ω

         So, R_T = 16 Ω

(b) the current through and voltage across each each resistor.

the total current is, i_T = V/R_T = 24 V/16 Ω = 1.5 A

Because R₁, R₂₃ and R₄₅ the series combination, then i₁ = i₂₃ = i₄₅ = i_T = 1.5 A

the voltage on R₁, R₂₃ and R₄₅ is

V₁ = i₁R₁ = 1.5 A x 4 Ω = 6 V
V₂₃ = i₂₃R₂₃ = 1.5 A x 4.8 Ω = 7.2 V
V₄₅ = i₄₅R₄₅ = 1.5 A x 7.2 Ω = 10.8 V

Because R₂ and R₃ the parallel combination, then V₂ = V₃ = V₂₃ = 7.2 V

and R₄ and R₅ the parallel combination, then V₄ = V₅= V₄₅ = 10.8 V

so that, current on the resistor R₂, R₃, R₄ and R₅ is

i₂ = V₂₃/R₂ = 7.2 V/8 Ω = 0.9 A
i₃ = V₂₃/R₃ = 7.2 V/12 Ω = 0.6 A
i₄ = V₄₅/R₄ = 10.8 V/6 Ω = 1.8 A
i₅ = V₄₅/R₅ = 10.8 V/24 Ω = 0.44 A

Problem #5
What is shown below is a series / parallel circuit. Calculate the total series / parallel resistance shown below, if the level is installed between points A and B. (The magnitude R₁ = 7 Ω, R₂ = 2.5 Ω, R₃ = 7.5 Ω, R₄ = 5 Ω, R₅ = 3 Ω and R₆ = 2 Ω)
 
Answer;
(a) if the level is installed between points A and B

STEP 1: resistor R₅ and R₆ in series, Fig.(a)
 

R₅₆ = R₅ + R₆ = 3 Ω + 2 Ω = 5 Ω

STEP 2: resistor R₅₆ and R₄ in parallel, Fig.(b).

1/R₄₅₆ = 1/R₄ + 1/R₅₆ = 1/5Ω + 1/5Ω = 2/5Ω

R₄₅₆ = 2.5 Ω

STEP 3: resistor R₄₅₆ and R₂ in series, Fig.(c).

R₂₄₅₆ = R₂ + R₄₅₆ = 2.5 Ω + 2.5 Ω = 5 Ω

STEP 4: resistor R₂₄₅₆ and R₃ in parallel, Fig.(d).

1/R₂₃₄₅₆ = 1/R₃ + 1/R₂₄₅₆ = 1/7.5 Ω + 1/5Ω = 12.5/37.5Ω

R₂₃₄₅₆ = 3 Ω

STEP 5: resistor R₂₃₄₅₆ and R₁ in seri, Fig.(e).

R_AB  = R₂₃₄₅₆ + R₁ = 3 Ω + 7 Ω = 10 Ω   

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