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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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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

Step-by-step explanation:

Cross sectional area of a wire is calculated as;


A = (\pi d^2)/(4)

Resistance of a wire is calculated as;


R = (\rho L)/(A) \\\\R = (4\rho L)/(\pi d^2) \\\\

Resistance in wire A;


R = (4\rho _AL_A)/(\pi d_A^2)

Resistance in wire B;


R = (4\rho _BL_B)/(\pi d_B^2)

Power delivered in wire;


P = (V^2)/(R)

Power delivered in wire A;


P = (V^2_A)/(R_A)

Power delivered in wire B;


P = (V^2_B)/(R_B)

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


P_A = (V^2_A)/(R_A) = (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 = (V^2_B)/(R_B) = (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;


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

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


(P_A)/(P_B) = ((V^2_A d^2_A)/(L_A) )*((L_B)/(V^2_B d^2_B))\\\\

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


(P_A)/(P_B) = (( d^2_A)/(L_A) )* ((L_B)/(d^2_B) )

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


(P_A)/(P_B) = (( (7d_B)^2)/(7L_B) )* ((L_B)/(d^2_B) )\\\\(P_A)/(P_B) = (49d_B^2)/(7L_B) *(L_B)/(d^2_B) \\\\(P_A)/(P_B) =(49)/(7) \\\\(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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