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:
RT = R1 + R2 + R3
RT = 3 Ω + 5 Ω + 4 Ω = 12 Ω
(b) the total current
i = V/RT = 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
PT = VTi = 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/RT = 1/R1 + 1/R2 + 1/R3
1/RT = 1/8 Ω + 1/6 Ω + 1/12 Ω = 18/48 Ω
RT = 2.67 Ω
(b) the total current
i = V/RT = 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
PT = VTi = 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,
Rp = R2R3/(R2 + R3) = (10 Ω)(15 Ω)/(10 Ω + 15 Ω) = 6 Ω
- R1 and Rp arranged in series, then
RT = R1 + Rp = 2 Ω + 6 Ω = 8 Ω
(b) the current through each resistor
the total current is, iT = V/RT = 24 V/8 Ω = 3 A
iT pass R1, then i1 = iT = 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
RT = R1 + R23 + R45 = 4 Ω + 4.8 Ω + 7.2 Ω
So, RT = 16 Ω
(b) the current through and voltage across each each resistor.
the total current is, iT = V/RT = 24 V/16 Ω = 1.5 A
Because R1, R23 and R45 the series combination, then i1 = i23 = i45 = iT = 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).
RAB = R23456 + R1 = 3 Ω + 7 Ω = 10 Ω
Post a Comment for "Resistors in Parallel and in Series Circuits Problems and Solutions"