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| learn:fundamentals:circuits:current:start [2016/07/22 21:37] – created Martha | learn:fundamentals:circuits:current:start [2022/10/04 11:38] (current) – external edit 127.0.0.1 | ||
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| ==== Answers ==== | ==== Answers ==== | ||
| 1. | 1. | ||
| - | * I = 3A. (The current leaving node b must equal the current entering node a.) | + | * a.) I = 3A. (The current leaving node b must equal the current entering node a.) |
| - | * I = -2A. (The current entering node b is 2A. Therefore, we must have 2A //leaving// node a. However, the sign convention associated with $I$ is that positive current is // | + | * b.) I = -2A. (The current entering node b is 2A. Therefore, we must have 2A //leaving// node a. However, the sign convention associated with $I$ is that positive current is // |
| - | * I = -3A. (The current leaving node b is equal to the current entering node a.) | + | * c.) I = -3A. (The current leaving node b is equal to the current entering node a.) |
| 2. | 2. | ||
| - | * 3A is entering node a. | + | * a.) 3A is entering node a. |
| - | * 2A is entering node b. (Since 2A is leaving node a.) | + | * b.) 2A is entering node b. (Since 2A is leaving node a.) |
| - | * 4A is entering node b. (Since // | + | * c.) 4A is entering node b. (Since // |
| {{tag> | {{tag> | ||

