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Here I know how we apply nodal analysis, but can somebody explain how can we calculate the current 'i' if there is no resistor in that particular wire?

I have shown my approach and working so far. I would really appreciate any guidance to steer me through the next steps to the solution.

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SamGibson
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somanshu
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    Thanks for the addition. Now, before your analysis starts: *Replace the wire labeled \$i\$ with a battery that has \$0\:\text{V}\$ across it. Now ground the bottom node (call it zero.) Swap the series \$8\:\text{V}\$ with the \$1\:\Omega\$ resistor (doesn't change anything important.) The resistor in series with the current source is irrelevant. Throw it out and replace it with a wire.* Pretty much all your nodes have voltages now. It's just a matter of KCL and you have your figure. – jonk Apr 21 '21 at 08:00
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    Thanks for improving your question, upvoted. Please add more to your question as you can but not just photo after photo of notes, it makes for a giant scroll-through to read and understand. Type in your workings where possible. Thanks. – TonyM Apr 21 '21 at 08:06

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Thanks for the addition. Just follow these steps:

schematic

simulate this circuit – Schematic created using CircuitLab

At this point, you've only one unknown node voltage, \$V_x\$, which is now easily computed using nodal analysis (or by simple inspection.) With \$V_x\$ in hand, you can work out all the currents in resistors \$R_1\$ through \$R_4\$.

(The current in \$R_5\$ is obviously already given. \$R_5\$ can be simply shorted out if you prefer, as its only impact is upon the voltage across the current source and no one cares what that is, anyway.)

With all the currents in hand, you can now use KCL on the top-right corner node to work out the current and its direction in \$V_2\$. And then use KCL on the node just above \$i\$ to work out the magnitude and direction of \$i\$.

jonk
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