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Three currents i1, i2 and i3 are meets at a node. If i1 = 10 sin(400t + 60°) A and i2 = 10 sin(400t - 60°) A then i3
- A0
- B10 sin 400t A
- C-10 sin 400t A
- D-5√3 sin 400t A
Answer not set
Discussion 1
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CCommunity
i1 = 10 sin(400t + 60°) A = 10∠60°
i2 = 10 sin(400t - 60°) A = 10∠-60°
Sum of three currents = 0
i1 + i2 + i3 = 0
⇒ 10∠60° + 10∠-60° + i3 = 0
⇒ 10 cos 60° + 10 sin 60° + 10 cos 60° - 10 sin 60° + i3 = 0
⇒ i3 = -10∠0°
⇒ i3 = -10 sin 400t A
i2 = 10 sin(400t - 60°) A = 10∠-60°
Sum of three currents = 0
i1 + i2 + i3 = 0
⇒ 10∠60° + 10∠-60° + i3 = 0
⇒ 10 cos 60° + 10 sin 60° + 10 cos 60° - 10 sin 60° + i3 = 0
⇒ i3 = -10∠0°
⇒ i3 = -10 sin 400t A