Norton: Two Loop Problem

To apply Norton's Theorem to the solution of the two loop problem, consider the current through resistor R₂ below. Replacing the remainder of the network by its Norton equivalent simplifies the determination of I₂ .

For R1 =Ω, R2 =Ω, R3 =Ω, 对于 R 1 = Ω，R 2 = Ω，R 3 = Ω， 中文译文中的待填/计算数值依次对应：1：r1 2：r2 3：r3。实际数值以上方原输入框为准。

and voltages V1 = V and V2 = V, 并且电压 V₁ = V 和 V₂ = V， 中文译文中的待填/计算数值依次对应：1：v1 2：v2。实际数值以上方原输入框为准。

the open circuit voltage is 开路电压是

= V

since R1 and R3 form a simple voltage divider. 由于R1和R3构成一个简单的电压分压器。

The Norton resistance is 诺顿电阻是

= Ω.

This gives a Norton current 这给出了诺顿电流

= A.

Finally the calculated current is 最后计算出的电流是

= A

Note: To avoid dealing with so many short circuits, any resistor with value zero will default to 1 when a voltage is changed. It can be changed back to a zero value if you wish to explore the effects of short circuits. Ohms and amperes are the default units, but if you put in resistor values in kilohms, then the currents will be milliamperes.