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)](https://img.qammunity.org/2021/formulas/physics/college/m0uk6ud1him8xy9nvl50spmfc5a24ozsug.png)
Resistance of a wire is calculated as;
![R = (\rho L)/(A) \\\\R = (4\rho L)/(\pi d^2) \\\\](https://img.qammunity.org/2021/formulas/physics/college/h1ip2olkq61w2xgkopv83ftixk22iw3djq.png)
Resistance in wire A;
![R = (4\rho _AL_A)/(\pi d_A^2)](https://img.qammunity.org/2021/formulas/physics/college/vvzzgznd8pv349ydsop1tkyf3xebve6pg5.png)
Resistance in wire B;
![R = (4\rho _BL_B)/(\pi d_B^2)](https://img.qammunity.org/2021/formulas/physics/college/2pioo4fe4f7iczn8f4qoxlm9c6ekl1afem.png)
Power delivered in wire;
![P = (V^2)/(R)](https://img.qammunity.org/2021/formulas/physics/middle-school/hkvkzxb1fu845dne22xvyraglrwqgg4qua.png)
Power delivered in wire A;
![P = (V^2_A)/(R_A)](https://img.qammunity.org/2021/formulas/physics/college/t0merg70nu3b9ptbluso9ujte9qurc42ia.png)
Power delivered in wire B;
![P = (V^2_B)/(R_B)](https://img.qammunity.org/2021/formulas/physics/college/c5x0ob1a8jatrjgfkg9jxh0pubkw0jg1jh.png)
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)](https://img.qammunity.org/2021/formulas/physics/college/snp2hzf4cn75j09z09jf6mbq95cq68xnt1.png)
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)](https://img.qammunity.org/2021/formulas/physics/college/czqqny8kz15591y7jp0gjef13t510f8xa9.png)
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))\\\\](https://img.qammunity.org/2021/formulas/physics/college/2a0eyn04j2z1prpvemez08dw2z7i141e08.png)
The wires are made of the same material,
![(P_A)/(P_B) = ((V^2_A d^2_A)/(L_A) )*((L_B)/(V^2_B d^2_B))\\\\](https://img.qammunity.org/2021/formulas/physics/college/eloeqm8v47lxu1ork3nqko4wdu1emx1j1r.png)
The wires are connected across the same potential;
![V_A = V_B](https://img.qammunity.org/2021/formulas/physics/college/1ilgajphhjog2js8kwuf7b8jq1dd12cobj.png)
![(P_A)/(P_B) = (( d^2_A)/(L_A) )* ((L_B)/(d^2_B) )](https://img.qammunity.org/2021/formulas/physics/college/gpew9ezkiqyg4hwtehz81r126g2c7kng4h.png)
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](https://img.qammunity.org/2021/formulas/physics/college/n002f74nke6qpq8dx41xtoq80bnbc9a90q.png)
Therefore, the ratio of the power delivered to A to power delivered to B is
7 : 1