회로 해석법⁕(ICA)

$$12 =2 I_1 + 2 (I_1 -I_2 ) + 1(I_1 -I_3 )$$ $$\to 5I_1 -2 I_2 -I_3 =12$$

$$0=2(I _{2} -I _{1} )+1I _{2} +1(I _{2} -I_3 )$$ $$\to -2I_1 +4I_2 -I_3 =0$$

$$0=1 (I_3 -I_2 ) + 2 I_3 + 1(I_3 -I_1 ) \to -I_1 -I_2 +4I_3 =0$$

$$\left|\begin{matrix}5,-2,-1\\-2,\ 4,-1\\-1,-1,4\end{matrix}\right|\left|\begin{matrix}I_1\\I_2\\I_3\end{matrix}\right|=\left|\begin{matrix}12\\0\\0\end{matrix}\right|$$

$$\left|\begin{matrix}a_1,b_1,c_1\\a_2,b_2,c_2\\a_3,b_3,c_3\end{matrix}\right|\left|\begin{matrix}x\\y\\z\end{matrix}\right|=\left|\begin{matrix}d_1\\d_2\\d_3\end{matrix}\right|$$

$$∴ c-d 사이의 전위차 = 1 [Ω] \times(I_2 -I_3 )$$ $$=1 [Ω] \times(2.12 -1.41 ) =0.71[V]$$ $$I_1 =3.53[A], I_2 =2.12[A], I_3 =1.41[A]$$