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You cannot remove dependent sources in applying superposition theorem. For removing voltage sources, you have to short them. For removing current source, open circuit them .Do nodal analysis for each step and see if you get the result.

Small hint:

Use superposition theorem to calculate \$V_1\$ and \$V_2\$ by adding \$V_1\$ and \$V_2\$ when current source is open circuit and when voltage source is short circuited while keeping the other source and the dependent source active. With that, you can calculate currents.

Big hint:

For example, removing the current source, I get these equations: $$\frac{V_1 - 3}{7} + \frac{V_1 - V_2}{15} = 0 \tag 1$$ Substituting current i, $$\frac{V_2}{5} + \frac{V_2 - V_1}{15} = 4 \frac{V_2}{5} \tag 2$$ From these,\$V_1\$ and \$V_2\$ can be calculated. Now, shorting voltage source and writing nodal equations. $$\frac{{V_1}^`}{7} + \frac{{V_1}^` - {V_2}^`}{15} = 2 \tag 3$$ $$\frac{{V_2}^`}{5} + \frac{{V_2}^` - {V_1}^`}{15} = 4 \frac{{V_2}^`}{5} \tag 4$$ From these, \$V_1\$ and \$V_2\$ and \${V_1}^`\$ and \${V_2}^`\$ can be calculated.

You cannot remove dependent sources in applying superposition theorem. For removing voltage sources, you have to short them. For removing current sources, open circuit them .Do nodal analysis for each step and see if you get the result.

Small hint:

Use superposition theorem to calculate \$V_1\$ and \$V_2\$ by adding \$V_1\$ and \$V_2\$ when current source is open circuit and when voltage source is short circuited while keeping the other source and the dependent source active. With that, you can calculate currents.

Big hint:

For example, removing the current source, I get these equations: $$\frac{V_1 - 3}{7} + \frac{V_1 - V_2}{15} = 0 \tag 1$$ Substituting current i, $$\frac{V_2}{5} + \frac{V_2 - V_1}{15} = 4 \frac{V_2}{5} \tag 2$$ From these,\$V_1\$ and \$V_2\$ can be calculated. Now, shorting voltage source and writing nodal equations. $$\frac{{V_1}^`}{7} + \frac{{V_1}^` - {V_2}^`}{15} = 2 \tag 3$$ $$\frac{{V_2}^`}{5} + \frac{{V_2}^` - {V_1}^`}{15} = 4 \frac{{V_2}^`}{5} \tag 4$$ From these, \$V_1\$ and \$V_2\$ and \${V_1}^`\$ and \${V_2}^`\$ can be calculated.

You cannot remove dependent sources in applying superposition theorem. For removing voltage sources, you have to short them. For removing current source, open circuit them .Do nodal analysis for each step and see if you get the result.

Small hint:

Use superposition theorem to calculate \$V_1\$ and \$V_2\$ by adding \$V_1\$ and \$V_2\$ when current source is open circuit and when voltage source is short circuited while keeping the other source and the dependent source active. With that, you can calculate currents.

Big hint:

For example, removing the current source, I get these equations: $$\frac{V_1 - 3}{7} + \frac{V_1 - V_2}{15} = 0 \tag 1$$ Substituting current i, $$\frac{V_2}{5} + \frac{V_2 - V_1}{15} = 4 \frac{V_2}{5} \tag 2$$ From these,\$V_1\$ and \$V_2\$ can be calculated. Now, shorting voltage source and writing nodal equations. $$\frac{V_1}{7} + \frac{V_1 - V_2}{15} = 2 \tag 3$$ $$\frac{V_2}{5} + \frac{V_2 - V_1}{15} = 4 \frac{V_2}{5} \tag 4$$ From these, \$V_1\$ and \$V_2\$ can be calculated.

You cannot remove dependent sources in applying superposition theorem. For removing voltage sources, you have to short them. For removing current source, open circuit them .Do nodal analysis for each step and see if you get the result.

Small hint:

Use superposition theorem to calculate \$V_1\$ and \$V_2\$ by adding \$V_1\$ and \$V_2\$ when current source is open circuit and when voltage source is short circuited while keeping the other source and the dependent source active. With that, you can calculate currents.

Big hint:

For example, removing the current source, I get these equations: $$\frac{V_1 - 3}{7} + \frac{V_1 - V_2}{15} = 0 \tag 1$$ Substituting current i, $$\frac{V_2}{5} + \frac{V_2 - V_1}{15} = 4 \frac{V_2}{5} \tag 2$$ From these,\$V_1\$ and \$V_2\$ can be calculated. Now, shorting voltage source and writing nodal equations. $$\frac{{V_1}^`}{7} + \frac{{V_1}^` - {V_2}^`}{15} = 2 \tag 3$$ $$\frac{{V_2}^`}{5} + \frac{{V_2}^` - {V_1}^`}{15} = 4 \frac{{V_2}^`}{5} \tag 4$$ From these, \$V_1\$ and \$V_2\$ and \${V_1}^`\$ and \${V_2}^`\$ can be calculated.

You cannot remove dependent sources in applying superposition theorem. For removing voltage sources, you have to short them. For removing current source, open circuit them .Do nodal analysis for each step and see if you get the result.

Small hint:

Use superposition theorem to calculate \$V_1\$ and \$V_2\$ by adding \$V_1\$ and \$V_2\$ when current source is open circuit and when voltage source is short circuited while keeping the other source and the dependent source active. With that, you can calculate currents.

Big hint:

For example, removing the current source, I get these equations: $$\frac{V_1 - 3}{7} + \frac{V_1 - V_2}{15} = 0 \tag 1$$ Substituting current i, $$\frac{V_2}{5} + \frac{V_2 - V_1}{15} = 4 \frac{V_2}{5} \tag 2$$ From these,\$V_1\$ and \$V_2\$ can be calculated. Now, shorting voltage source and writing nodal equations. $$\frac{V_1}{7} + \frac{V_1 - V_2}{15} = 2 \tag 1$$ $$\frac{V_2}{5} + \frac{V_2 - V_1}{15} = 4 \frac{V_2}{5} \tag 2$$ From these, \$V_1\$ and \$V_2\$ can be calculated.

You cannot remove dependent sources in applying superposition theorem. For removing voltage sources, you have to short them. For removing current source, open circuit them .Do nodal analysis for each step and see if you get the result.

Small hint:

Use superposition theorem to calculate \$V_1\$ and \$V_2\$ by adding \$V_1\$ and \$V_2\$ when current source is open circuit and when voltage source is short circuited while keeping the other source and the dependent source active. With that, you can calculate currents.

Big hint:

For example, removing the current source, I get these equations: $$\frac{V_1 - 3}{7} + \frac{V_1 - V_2}{15} = 0 \tag 1$$ Substituting current i, $$\frac{V_2}{5} + \frac{V_2 - V_1}{15} = 4 \frac{V_2}{5} \tag 2$$ From these,\$V_1\$ and \$V_2\$ can be calculated. Now, shorting voltage source and writing nodal equations. $$\frac{V_1}{7} + \frac{V_1 - V_2}{15} = 2 \tag 3$$ $$\frac{V_2}{5} + \frac{V_2 - V_1}{15} = 4 \frac{V_2}{5} \tag 4$$ From these, \$V_1\$ and \$V_2\$ can be calculated.