[ V_source = 2 \times I_x = 2 \times 0.5 = 1V ] Then ( V_out = V_source ) across the resistor? Actually if ( 4\Omega ) is in series, same current ( I_x ) flows through it: [ V_out = I_x \cdot 4\Omega = 0.5 \cdot 4 = 2V ] But the dependent source adds voltage. Need a full loop: Loop equation: ( -2I_x + V_out = 0 ) (if only resistor and source) → ( V_out = 2I_x = 1V ) ✅ Answer depends on topology. Typical answer: ( V_out = 1V )
: The sum of currents entering a node equals the sum of currents leaving it ( Second Law (KVL) elektroteknika 1 ushtrime full
Good luck with your studies in Elektroteknika! ⚡📐 [ V_source = 2 \times I_x = 2 \times 0
Gjatë zgjidhjes së ushtrimeve "full", fokusohuni në këto pika kryesore: Elektroteknika PDF - Scribd Typical answer: ( V_out = 1V ) :