phy electric current 2條

Q1.
R = k L/A, for k = resistivity, L = length of wire, A = cross-sectional area
P = I^2 R/2 = I^2 (KL/A)/2, for P = power dissipated, I = current
L_X = 3L_Y,
For Volume constant, A_X L_X = A_Y L_Y, i.e. 3A_X = A_Y
P_X : P_Y
= (I_X)^2 (KL_X /A_X)/2 : (I_Y)^2 (KL_Y /A_Y)/2
= 1^2 (3L_Y/A_X) : 2^2 (L_Y/3A_X)
= 9:4
Ans: D

Q2.
For V_T = constant, I_T = constant
Since R_2, R_3 are in parallel, they share the same p.d.
I_2R_2 = V and I_3R_3 = V
I_2 : I_3 = R_3 : R_2
By P = IV/2,
P_2:P_3 = 1:3
I_2 : I_3 = 1:3
R_3 : R_2 = 1:3
For R_3 = R,
R_2 = 3R
Combining R_C of R_2 and R_3 = 1/(1/R + 1/3R) = 3R/4
Since L_1 and L_C are in series, they share the same current
I = V_1/R_1 and I = V_C/R_C
V_1:V_C = R_1:R_C
By P = IV/2,
P_1 : P_C = 1:4
V_1 : V_C = 1:4
R_1 : R_C = 1:4
R_1 : 3R/4 = 1:4
R_1 = 3R/16
Ans:A

1. Let R1 and R2 be the resistances of metal rods X and Y respectively
L1 and L2 be the lengths of X and Y respectively
A1 and A2 be the cross-sectional areas of X and Y respectively

Given that: L1 = 3(L2)
and, mass = d.(A1).(L1) = d.(A2).(L2)
where d is the density of the material
i.e. A2/A1 = L1/L2 = 3

Resistance of X, R1 = p(L1)/A1
Resistance of Y, R2 = p(L2)/A2
where p is the resistivity of the material
Hence, R1/R2 = (L1/L2).(A2/A1) = 3 x 3 = 9

Power dissipated on X, P1 = (1^2).R1
Power dissipated on Y, P2 = (2^2).R2
Hence, P1/P2 = R1/4(R2) = (1/4).(R1/R2) = 9/4

2. Let the resistances of L1 and L2 be R1 and R2 respectively

Power consumed by L2, P2 = V^2/R2, V is the voltage across L2
Power consumed by L3, P3 = V^2/R
hence, P2/P3 = 1/3 = R/R2
i.e. R2 = 3R

Let I1, I2 and I3 be the current through L1, L2 and L3 respectively
Since power consumed by L1 and L2 are equal
(I1)^2.(R1) = (I2)^2.R2
i.e. (I1)^2.(R1) = (I2)^2.(3R)
R1 = (I2/I1)^2.(3R) ---------------------------- (1)

But P2 = (I2)^2.R2 = (I2)^2.(3R)
P3 = (I3)^2.(R) = (I1-I2)^2.R
P2/P3 = 1/3 = 3.[I2/(I1-I2)]^2
i.e. 1/9 = [I2/(I1-I2)]^2
I2/(I1-I2) = 1/3
3.(I2) = I1 - I2
4(I2) = I1
I2/I1 = 1/4

Substitute into (1),
R1 = (1/4)^2.3R = 3R/16