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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: More power is delivered to conductor 2, which has twice the resistance of conductor 1.

Step by step solution

01

Write the givens and the equation for the total resistance

We are given that the resistance of one conductor is twice that of the other. Let's denote the resistance of the first conductor as R1 and the resistance of the second conductor as R2. We have the relationship: R2 = 2 * R1 Since both conductors are connected to the same emf device, they are in series, and the total resistance of the circuit is: R_total = R1 + R2
02

Express R2 in terms of R1

We will now express R2 in terms of R1 using the given relationship: R2 = 2 * R1 We can substitute this expression into the equation for the total resistance: R_total = R1 + 2 * R1 = 3 * R1
03

Calculate the current through the circuit

Using Ohm's law, we can calculate the current through the circuit: I = E / R_total We will substitute the expression for R_total into the equation: I = E / (3 * R1)
04

Calculate the power delivered to each conductor

We can now calculate the power delivered to each conductor using the power formula for a conductor: P1 = I^2 * R1 P2 = I^2 * R2 We will substitute the expressions for I and R2 into these equations: P1 = (E / (3 * R1))^2 * R1 P2 = (E / (3 * R1))^2 * (2 * R1)
05

Compare P1 and P2

We can now compare the power delivered to each conductor: P1 = (E^2 / (9 * R1^2)) * R1 P2 = (E^2 / (9 * R1^2)) * (2 * R1) We can clearly see that P2 has an extra factor of 2 in its equation, and since the rest of the terms are the same for both equations, the power delivered to conductor 2 (P2) is twice that of conductor 1 (P1). Hence, more power is delivered to the conductor with twice the resistance, which is conductor 2.

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