物理問題 電學2 求教~~急(10點)

(ii B) In applying the KVL, we should also take the polarity of each component's voltage into account, also their corresponding direction. Considering loop 2 as follows:

圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Nov11/Crazyelec2.jpg


With the polarities (+/-) mentioned as above, we have:

ε, 20Ω and 50Ω resistors' voltage are in the same direction, in opposite to the 5 V cell.

So summing up, we have:

ε + 20I2 + 50I3 = 5 ... (*)

giving

ε - 5 = -20I2 - 20I3

So if you prefer to have 5 - ε, then it would become:

5 - ε = 20I2 + 50I3 which can also be derived from (*)

(iii) Taking the middle branch:


圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Nov11/Crazyelec1.jpg


I3 indeed flows from Y to Z and then Z to X since it is engative, hence:

VYZ = 50 x 0.22 = 11 V with Y being higher voltage than Z

VZX = 5 V with Z being of higher voltage than X

Hence VYX = 16 V with Y being of higher voltage than X.

(iv) As derived in (iii).

(v) Since VZX = 5 V with Z being of higher voltage than X, then

VYZ = 5 V with Z being of higher voltage than Y so that X and Y will be of the same potential

Hence I3 should flow from Z to Y with value = 5/50 = 0.1 A.

(vi) Since I3 flows from Z to Y as described in (v), it is equivalent to flowing from X to Y.

(iib) Yes, you could equally write: 5 – ε = 50(I3) + 20(I2)The rule is that if you go along with the direction of the loop given on the diagram, any current or voltage in the same direction of the loop is taken as +ve and against its direction is taken as –ve. Hence, you would have +ve for ε, and –ve for 5v, I2 and I3. (iii) Since I3 is –ve, the current in fact flows from Y to X. That is Y is at a higher potential than X (because current flows from high to low potentials). Hence, voltage drop from Y to X is [(0.22 x 50) + 5] v = 16 v (iv) See answer given to (iii) above. (v) If the voltage across XY is zero, this indicates that when you go from X to Y, the voltage rise of 5 v of the battery is balanced by the voltage drop across the 50-ohm resistor. Hnece, 5 = (I3’).50where I3’ is the new currenti.e. I3’ = 5/50 A = 0.1 A (vi) Yes. Current flows from X to Y gives a potential rise across the battery and a potential drop across the resistor. This satisfies the condition stated in (v) above. If current flows from Y to X, potential drops would result across both the battery and resistor.