Two conductors made of the same material are connected across the same potential difference. Conductor A has seven times the diameter and se

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Two conductors made of the same material are connected across the same potential difference. Conductor A has seven times the diameter and seven times the length of conductor B. What is the ratio of the power delivere

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Philomena 3 years 2021-08-22T16:28:26+00:00 1 Answers 7 views 0

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  1. RobertKer
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    2021-08-22T16:30:09+00:00
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    Complete question:

    Two conductors made of the same material are connected across the same potential difference. Conductor A has seven times the diameter and seven times the length of conductor B. What is the ratio of the power delivered to A to power delivered to B.

    Answer:

    The ratio of the power delivered to A to power delivered to B is 7 : 1

    Explanation:

    Cross sectional area of a wire is calculated as;

    A = \frac{\pi d^2}{4}

    Resistance of a wire is calculated as;

    R = \frac{\rho L}{A} \\\\R = \frac{4\rho L}{\pi d^2} \\\\

    Resistance in wire A;

    R = \frac{4\rho _AL_A}{\pi d_A^2}

    Resistance in wire B;

    R = \frac{4\rho _BL_B}{\pi d_B^2}

    Power delivered in wire;

    P = \frac{V^2}{R}

    Power delivered in wire A;

    P = \frac{V^2_A}{R_A}

    Power delivered in wire B;

    P = \frac{V^2_B}{R_B}

    Substitute in the value of R in Power delivered in wire A;

    P_A = \frac{V^2_A}{R_A} = \frac{V^2_A \pi d^2_A}{4 \rho_A L_A}

    Substitute in the value of R in Power delivered in wire B;

    P_B = \frac{V^2_B}{R_B} = \frac{V^2_B \pi d^2_B}{4 \rho_B L_B}

    Take the ratio of power delivered to A to power delivered to B;

    \frac{P_A}{P_B} = (\frac{V^2_A \pi d^2_A}{4\rho_AL_A} ) *(\frac{4\rho_BL_B}{V^2_B \pi d^2_B})\\\\ \frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{\rho_AL_A} )*(\frac{\rho_BL_B}{V^2_B d^2_B})\\\\

    The wires are made of the same material, \rho _A = \rho_B

    \frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{L_A} )*(\frac{L_B}{V^2_B d^2_B})\\\\

    The wires are connected across the same potential; V_A = V_B

    \frac{P_A}{P_B} = (\frac{ d^2_A}{L_A} )* (\frac{L_B}{d^2_B} )

    wire A has seven times the diameter and seven times the length of wire B;

    \frac{P_A}{P_B} = (\frac{ (7d_B)^2}{7L_B} )* (\frac{L_B}{d^2_B} )\\\\\frac{P_A}{P_B} = \frac{49d_B^2}{7L_B} *\frac{L_B}{d^2_B} \\\\\frac{P_A}{P_B} =\frac{49}{7} \\\\\frac{P_A}{P_B} = 7\\\\P_A : P_B = 7:1

    Therefore, the ratio of the power delivered to A to power delivered to B is

    7 : 1

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