Physics, 24.03.2020 17:04 megamorph

A single resistor is connected across the terminals of a battery. Which one or more of the following changes in voltage and current leaves unchanged the electric power dissipated in the resistor? (Show work) (A) Doubling the voltage and increasing the resistance by a factor of four (B) Doubling the voltage and the current (C) Doubling the voltage and reducing the resistance by a factor of four (D) Doubling the current and reducing the resistance by a factor of four (a) All (b) A, C, D only (c) B, C, D only (d) A and D only

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Answer from: sa9334

Answer is option (d) - (A and D only).

(A) Doubling the voltage and increasing the resistance by a factor of four.

(D) Doubling the current and reducing the resistance by a factor of four

Explanation:

According to Ohm's law, the electrical current, flowing through a conductor having a fixed resistance, is directly proportional to the voltage, applied across it and is inversely proportional to the resistance, i.e., V=IR .

The rate at which energy is produced or absorbed within an electrical circuit is known as the electrical power, . In a circuit, the energy source such as a voltage from the battery delivers power to the connected load such as a single resistor, light bulbs, etc. These loads absorb electrical power and convert it into heat or light or both.

The equations for calculating power are;

P=VI ......... (1)

$P=\frac{V^{2}}{R}$ .......... (2)

$P=I^{2} R$ .......... (3)

(i) In statement (A) of the question, it is given that the voltage is doubled, i.e., now the voltage is and the resistance is increased by a factor of four, i.e., now the resistance is .

Therefore, by applying the equation (2),

$P= \frac{(2V)^{2}}{4R} = \frac{4V^{2}}{4R} =\frac{V^{2}}{R}$ ......... (4)

From equation (4), it is clear that the power remains unchanged. So, the changes in voltage and current do not change the electric power dissipated in the resistor.

(ii) In statement (B) of the question, it is given that both the voltage and the current are doubled, i.e., now the voltage is and the current is .

Therefore, by applying the equation (1),

P=(2V)(2I)=4VI .............. (5)

From equation (5), it is clear that power is increased by a factor of four. So, the changes in voltage and current changes the electric power dissipated in the resistor.

(iii) In statement (C) of the question, it is given that the voltage is doubled, i.e., now the voltage is and the resistance is decreased by a factor of four, i.e., now the resistance is $\frac{R}{4}$ .

Therefore, by applying the equation (2),

$P= \frac{(2V)^{2}}{\frac{R}{4} } = \frac{4V^{2}}{\frac{R}{4}} =\frac{16V^{2}}{R}$ ......... (6)

From equation (6), it is clear that power is increased by a factor of sixteen. So, the changes in voltage and current changes the electric power dissipated in the resistor.

(iv) In statement (D) of the question, it is given that the current is doubled, i.e., now the current is and the resistance is decreased by a factor of four, i.e., now the resistance is $\frac{R}{4}$ .

Therefore, by applying the equation (3),

$P=(2I)^{2} *(\frac{R}{4} )=(4I^{2} )*(\frac{R}{4} )=I^{2} R$ ........ (7)

From equation (7), it is clear that the power remains unchanged. So, the changes in voltage and current do not change the electric power dissipated in the resistor.

Thus, the statements (A) and (D) are the correct answer.

Answer from: Quest

Need more information like a picture

Answer from: Quest

The independent variable in this experiment is the angle at which the light from the ray box strikes the mirror.hope this .

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