物理:求安培計及電阻器的電阻

Just treat each ammeter as a resistor with resistance Ra.
Since current flows through M1 is 4A and that flows through M2 is 2A, the current flows through the lower-right resistor is (4 - 2) A = 2 A

Apply Ohm's Law to the loop from battery through M1, the lower-right resistor and back to battery, we have,
10 = 4R + 2(Ra)
i.e. R = (5 - Ra)/2 ------------ (1)

Consider the path from battery to M1, M2, M3 and then back to battery, using Ohm's Law,
10 = 4(Ra) + 2(Ra) + 4(Ra)
i.e. 10 = 10(Ra)
Ra = 1 ohm

From (1), R = (5 - 1)/2 ohms = 2 ohms
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This is in fact a Wheatstone Bridge circuit, not a simple parallel circuit.


2015-04-03 16:41:03 補充：
sorry, I made a mistake. The equation should be
10 = 4(Ra) + 2R
i.e. R = [10 - 4(Ra)]/2
We then found Ra = 1 ohm
R = (10 - 4/2 ohms = 3 ohms

2015-04-03 16:45:51 補充：
Q: 為什麼有 10 = 4(Ra) + 2(Ra) + 4(Ra)
A: Just follow the path from +ve pole of battery to M1, M2, M3 and then back to -ve pole of battery.
Voltage acroos M1 = 4(Ra), across M2 = 2(Ra), across M3 = 4(Ra)

2015-04-03 16:46:44 補充：
(cont'd)...
Since battery voltage = voltage across M1 + voltage across M2 + voltage across M3
i.e. 10 = 4(Ra) + 2(Ra) + 4(Ra)

2015-04-03 16:51:44 補充：
You could refer to the following link for explanation of Wheastone Bridge (惠斯登電橋):

http://zh.wikipedia.org/wiki/%E6%83%A0%E6%96%AF%E7%99%BB%E9%9B%BB%E6%A9%8B

圖請見:
photo.pchome.com.tw/forever2015/142802815053