GATE EE 2016 Set 1 — Question 65
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Question
In the circuit shown below, the node voltage is ___________ V.
Correct answer
11.25 to 11.5
Solution
To find the node voltage , we apply Kirchhoff's Current Law (KCL) at node .
The total resistance in the right branch is . The voltage across this branch is . Thus:
Multiply the entire equation by 5:
1.Identify the branches connected to node :
- Leftmost branch: A resistor connected to ground. Current leaving node is .
- Second branch: A current source pointing upwards. Current entering node is .
- Middle branch: A resistor in series with a dependent voltage source (polarity: at top, at bottom). Current leaving node downwards is .
- Right branch: A horizontal resistor in series with a vertical resistor and a voltage source (polarity: at top, at bottom). The current through this entire branch is .
The total resistance in the right branch is . The voltage across this branch is . Thus:
3.Apply KCL at node :
Sum of currents leaving node Multiply the entire equation by 5:
4.Substitute into the KCL equation:
The value of is approximately , which falls within the specified range of to .Continue learning with Success Tracker
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