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For the network shown below, the Thevenin’s voltage Vab is

For the network shown below, the Thevenin’s voltage Vab is Correct Answer -1.5 V

Thevenin’s voltage is the open circuit voltage across the terminals a and b.

By using source transformation,

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By using source transformation once again,

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By applying KVL

30 + 15 I + 5I + 8 = 0

⇒ I = -1.9 A

By applying KVL

30 + 15I + Vab = 0

⇒ Vab = -15I – 30 = -15(-1.9) - 30

= -1.5 V

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