2條計電壓問題,不會做

1. Follow the path from the +ve to -ve terminals of the battery round the loop. We have,

20 = V1 + 2 + V2 + V1
i.e. 20 -2 = 2(V1) + V2
18 = 2(V1) + V2 ---------- (1)

Follow the path of the half loop to the right of 10 v, we have,
10 = 2 + V2
i.e. V2 = 8 v

Hence, substitute V2 into (1),
18 = 2(V1) + 8
V1 = 5 v

2. Consider the left-most loop, the current returning to the -ve pole of the battery is 12.6 mA (equals to that coming out from the +ve pole). Since the current through R1 is I1, at the junction below R1,
12.6 = I1 + 8.5
i.e. I1 = (12.6 - 8.5) mA = 4.1 mA

Consider the right-most loop, the current through R3 is 4 mA. Hence, at the junction below R2,
4 + I2 = 8.5
I2 = (8.5 - 4 ) mA = 4.5 mA



2014-09-11 23:39:24 補充：
sorry, I made a mistake in Q1.
Equation (1) should be: 20 = V1 + 2 + V2 + 1
i.e. 17 = V1 + V2
Hence, V1 = 17 - V2 = (17 - 8) v = 9 v

2+V2=10
V2=8V

V1+10+1=20
V1=9V



I2+0.004=0.0085
I2=0.0045=4.5mA

I1+0.0085=0.0126
I1=0.0041=4.1mA