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Two conductors of the same length and radius are connected to the same emf device. If the resistance of one is twice that of the other, to which conductor is more power delivered?

Short Answer

Expert verified
Answer: The conductor with lower resistance has more power delivered to it.

Step by step solution

01

Recall Ohm's Law and the expression for power

According to Ohm's Law, the electric current (I) flowing through a conductor is inversely proportional to its resistance (R) and directly proportional to the voltage (V) across the conductor: I = V / R Also, the power (P) delivered to a conductor is given by the expression: P = I^2 * R
02

Express the currents in terms of resistance

Let's denote the resistance of the first conductor as R1 and the resistance of the second conductor as R2. Given that the resistance of one conductor is twice that of the other: R2 = 2 * R1 Since both conductors are connected to the same emf device, the voltage across both conductors is the same. Let's denote the voltage as V. Using Ohm's Law, we can express the current flowing through each conductor in terms of its resistance: I1 = V / R1 I2 = V / (2 * R1)
03

Calculate the power delivered to each conductor

Now we can calculate the power delivered to each conductor using the expression for power: P1 = I1^2 * R1 P2 = I2^2 * R2 Substitute the expressions for I1 and I2: P1 = (V / R1)^2 * R1 P2 = (V / (2 * R1))^2 * (2 * R1)
04

Compare the power delivered to each conductor

To compare the power delivered to each conductor, we can simplify the expressions for P1 and P2: P1 = V^2 / R1 P2 = V^2 / (4 * R1) * (2 * R1) = V^2 / (2 * R1) Since P1 = V^2 / R1 and P2 = V^2 / (2 * R1), it's clear that P1 > P2. Therefore, more power is delivered to the conductor with the lower resistance (R1).

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