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Meenie Leis
Meenie Leis
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Because the current through \$ 1\Omega\$ is not \$2-I_s\$ as you have assumed. It is actually = \$2-(I_s+I_{2\Omega})\$, where \$I_{2\Omega} = 10/2 = 5A\$. You didn't see that there are actually 4 branches at the upper node where you split the current using KCL. Not 3 nodesbranches. \$2\Omega\$ is the branch you missed. I think \$2\Omega\$ was deliberately drawn like that to trick students.

Because the current through \$ 1\Omega\$ is not \$2-I_s\$ as you have assumed. It is actually = \$2-(I_s+I_{2\Omega})\$, where \$I_{2\Omega} = 10/2 = 5A\$. You didn't see that there are actually 4 branches at the upper node where you split the current using KCL. Not 3 nodes. \$2\Omega\$ is the branch you missed. I think \$2\Omega\$ was deliberately drawn like that to trick students.

Because the current through \$ 1\Omega\$ is not \$2-I_s\$ as you have assumed. It is actually = \$2-(I_s+I_{2\Omega})\$, where \$I_{2\Omega} = 10/2 = 5A\$. You didn't see that there are actually 4 branches at the upper node where you split the current using KCL. Not 3 branches. \$2\Omega\$ is the branch you missed. I think \$2\Omega\$ was deliberately drawn like that to trick students.

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Meenie Leis
Meenie Leis
  • 2.8k
  • 13
  • 31

Because the current through \$ 1\Omega\$ is not \$2-I_s\$ as you have assumed. It is actually = \$2-(I_s+I_{2\Omega})\$, where \$I_{2\Omega} = 10/2 = 5A\$. You didn't see that there are actually 4 branches at the upper node where you split the current using KCL. Not 3 nodes. \$2\Omega\$ is the branch you missed. I think \$2\Omega\$ was deliberately drawn like that to trick students.