Quiz 9 Flashcards
How to Calculate Power in DC Series Circuits (17 cards)
Power delivered to a DC circuit is the measure of the amount of ? the circuit per unit of time.
a. current flowing through
b. horsepower
c. voltage drop in
d. work done by
d. work done by
When reviewing the power consumed by the circuit, what are the two possible ways in which the power is consumed?
a. Current flow / heat
b. Heat / useful work
c. Photo-effect / useful work
d. Useful work / voltage drop
b. Heat / useful work
The sum of the power dissipated by each resistor in a series circuit is equal to the ? .
a. total current in the series branch
b. total power
c. total resistance
d. voltage times the resistance
b. total power
Decreasing the voltage applied to a standard wattage lamp below that recommended by the manufacturer would result in a decrease in the amount of power consumed by the lamp.
True or False?
True.
What would happen to a resistor if it was subjected to voltages or currents which caused it to dissipate more power than the value for which it was rated?
a. The resistance value would increase.
b. The resistance value would decrease.
c. It would be damaged.
d. None of the above.
c. It would be damaged.
Which formula would be used to find the maximum current that can be carried by a resistor if the resistor’s resistance and wattage rating are known?
a. Iₜ = Rₜ × Pₜ
b. Iₜ = Pₜ / Rₜ
c. Iₜ = √(Rₜ / Pₜ)
d. Iₜ = √(Pₜ / Rₜ)
d. Iₜ = √(Pₜ / Rₜ)
The total power requirement in a series circuit is the sum of the powers of the individual components in that circuit.
True or False?
True.
As the current through a series circuit increases, the total power dissipated in each component decreases.
True or False?
False.
Which is the formula used to determine the total power in a series circuit, when the total resistance and current of that circuit are known?
a. Pₜ = Rₜ × Eₜ
b. Pₜ = Iₜ × Rₜ²
c. Pₜ = √(Rₜ × Iₜ)
d. Pₜ = Rₜ × Iₜ²
d. Pₜ = Rₜ × Iₜ²
If the current through a resistor is reduced by one-half, the power being dissipated by the resistor will be ? of the original power.
Rₜ = 5 Ω
Iₜ = 5 A
a. 1/8
b. 1/4
c. 1/2
d. 3/4
b. 1/4
When the current in the circuit is cut half, the voltage at the source is also cut in half. This results in 1/4 the original power.
Find the total power dissipated in the circuit.
Eₜ = 36 V
Rₜ = 72 Ω
18 W
Iₜ = Eₜ / Rₜ
= 36 / 72
= 0.5 A
Pₜ = Eₜ × Iₜ
= 36 × 0.5
= 18 W
What would be the minimum wattage rating value for resistor R2 in the circuit?
Eₜ = 180 V
R₁ = 27 Ω
R₂ = 18 Ω
288 W
Rₜ = R₁ + R₂
= 27 + 18
= 45 Ω
Iₜ = Eₜ / Rₜ
= 180 / 45
= 4 A
Pᴿ² = Iᵣ₂² × R₂
= 4² × 18
= 288 W
Given the circuit and parameters shown, find the total power for the circuit by determining the power consumed by each resistor. Be prepared to show your work in class.
Eᵣ₁ = 20 V
Eᵣ₂ = 50 V
Eᵣ₃ = 35 V
Iₜ = 2.5 A
262.5
Iᵣ₁ = Iₜ
=2.5 A
Pᵣ₁ = Eᵣ₁ × Iᵣ₁
= 20 × 2.5
= 50 W
Iᵣ₂ = Iₜ
= 2.5 A
Pᵣ₂ = Eᵣ₂ × Iᵣ₂
= 50 × 2.5
= 125 A
Iᵣ₃ = Iₜ
= 2.5 A
Pᵣ₃ = Eᵣ₃ × Iᵣ₃
= 35 × 2.5
= 87.5
Pₜ = Pᵣ₁ + Pᵣ₂ + Pᵣ₃ + Pᵣ₄ + Pᵣ₅ + ….Pₙ
= 50 + 125 + 87.5
= 262.5 W
If a 120-volt, 50 watt lamp is connected in series with a 120-volt, 75-watt lamp, and the series circuit is supplied with a 240-volt source, the lamps will operate at their rated power outputs.
True or False?
False
The lamps will not operate at their specified power outputs. The wattage rating of a lamp is based on both the voltage applied to that lamp and the current through the lamp. In a series circuit, the current through all components of the circuit must be the same, and since the lamps specified have the same voltage ratings, their current ratings must be different in order to have different power ratings. The lamps will operate at other than their normal ratings.
In the circuit shown, the four resistors dissipate a total of 150 watts and each dissipates an equal amount of power. Solve for the following:
What is the resistance of R₁?
What is the power dissipated by R₁?
Eₜ = 24 V
Rₜ = 960 mΩ
Pᵣ₁ = 37.5 W
Pₜ = 150 W
Iₜ = Pₜ / Eₜ
= 150 / 24
= 6.25 A
Rₜ = Eₜ / Iₜ
= 24 / 6.25
= 3.84 Ω
R₁ = Rₜ / 4
= 3.84 / 4
= 0.96 Ω
Iᵣ₁ = Iₜ
= 6.25 A
Pᵣ₁ = Iᵣ₁² × R₁
= 6.25² × 0.96
= 37.5 W
In the circuit, the lamp is rated at 4 volts and 0.5 watts. What size resistor must R1 be in order to supply the correct current to this lamp?
Eₜ = 9 V
P₁ = 0.5 W
Eᵣ₂ = 4 V
40 Ω
Iₗ₁ = Pₗ₁ / Eₗ₁
= 0.5 / 4
= 0.125 A
Eᵣ₁ = Eₜ - Eᵣ₂
= 9 - 4
= 5 V
R₁ = Eᵣ₁ / Iᵣ₁
= 5 / 0.125
= 40 Ω
Find the total power in the circuit shown.
Eₜ = 12 V
R₁ = 75 Ω
R₂ = 150 Ω
R₃ = 300 Ω
Pₜ = 274.29 mW
Rₜ = R₁ + R₂ + R₃
= 75 + 150 + 300
= 525 Ω
Iₜ = Eₜ / Rₜ
= 12 / 525
= 0.022857
Pₜ = Rₜ × Iₜ²
= 525 / 0.022857²
= 0.2742857 W or 274.29 mW
