2.23 - resistor). Hence, . 2 + V 1 4 V 1 2 = V 1 = . 8 V ....

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Problem 2.23 Determine the amount of power supplied by the independent current source in the circuit of Fig. P2.23. V 1 I 2 Ω 2 Ω 0.2 A 4 V 1 + _ Figure P2.23: Circuit for Problem 2.23. Solution: KCL at top node gives 0 . 2 + V 1 4 I = 0 Also I = V 1 / 2 (for the 2-
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Unformatted text preview: resistor). Hence, . 2 + V 1 4 V 1 2 = V 1 = . 8 V . The voltage across the 0.2-A source is 2 V 1 = 1 . 6 V. Power dissipated is P = VI = 1 . 6 . 2 = . 32 W ....
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