Evaluate functions for k(x) =6x+100
1 answer:
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<h2>Answer:</h2>
The power consumed in the lamp marked 100W - 110V is 15.68W
The power consumed in the lamp marked 100W - 220V is 62.73W
<h2>Step-by-step explanation:</h2>
Given:
<em>First lamp rating</em>
Power (P) = 100W
Voltage (V) = 110V
<em>Second lamp rating</em>
Power (P) = 100W
Voltage (V) = 220V
<em>Source</em>
Voltage = 220V
i. <u>Get the resistance of each lamp</u>.
Remember that power (P) of each of the lamps is given by the quotient of the square of their voltage ratings (V) and their resistances (R). i.e
P =
<em>Make R subject of the formula</em>
⇒ R = ------------------(i)
<em />
<em>For first lamp, let the resistance be R₁. Now substitute R = R₁, V = 110V and P = 100W into equation (i)</em>
R₁ =
R₁ = 121Ω
<em />
<em>For second lamp, let the resistance be R₂. Now substitute R = R₂, V = 220V and P = 100W into equation (i)</em>
R₂ =
R₂ = 484Ω
<em />
<em />
ii.<u> Get the equivalent resistance of the resistances of the lamps.</u>
Since the lamps are connected in series, their equivalent resistance (R) is the sum of their individual resistances. i.e
R = R₁ + R₂
R = 121 + 484
R = 605Ω
iii. <u>Get the current flowing through each of the lamps. </u>
Since the lamps are connected in series, then the same current flows through them. This current (I) is produced by the source voltage (V = 220V) of the line and their equivalent resistance (R = 605Ω). i.e
V = IR [From Ohm's law]
I =
I =
I = 0.36A
iv. <u>Get the power consumed by each lamp.</u>
From Ohm's law, the power consumed is given by;
P = I²R
Where;
I = current flowing through the lamp
R = resistance of the lamp.
<em>For the first lamp, power consumed is given by;</em>
P = I²R [Where I = 0.36 and R = 121Ω]
P = (0.36)² x 121
P = 15.68W
<em>For the second lamp, power consumed is given by;</em>
P = I²R [Where I = 0.36 and R = 484Ω]
P = (0.36)² x 484
P = 62.73W
Therefore;
The power consumed in the lamp marked 100W - 110V is 15.68W
The power consumed in the lamp marked 100W - 220V is 62.73W