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Determine if the circuit is valid or invalid.
Please note which node I used as the ground node and which node I have labeled [itex]V_{1}[/itex]. By the node voltage method
[itex]\frac{V_{1}}{8} + 10 - 2 + \frac{V_{1} - 10}{5} = 0[/itex]
[itex]\frac{13V_{1}}{40} + 6 = 0 [/itex]
[itex]\frac{13V_{1}}{40} = -6[/itex]
[itex]V_{1} = -\frac{240}{13} V[/itex]
[itex]I_{1} = \frac{V_{1}}{8} = -\frac{240}{13*8} = -\frac{30}{13} A[/itex]
[itex]I_{2} = \frac{V_{1}-10}{5} = \frac{\frac{-240}{13}-10}{5} = -\frac{74}{13}[/itex]
[itex]I_{3} = 10 + I_{2} = 10 + {-\frac{74}{13}} = \frac{56}{13}[/itex]
[itex]V = IR[/itex]
[itex]P = IV[/itex]
[itex]P = I^{2}R[/itex]
[itex]P_{8Ω} = I^{2}R = (-\frac{30}{13})^{2}8 = \frac{7200}{169}[/itex]
[itex]P_{5Ω} = I^{2}R = (-\frac{74}{13})^{2}5 = \frac{27380}{169}[/itex]
[itex]P_{2Ω} = I^{2}R = (\frac{56}{13})^{2}2 = \frac{6272}{169}[/itex]
[itex]P_{10V} = IV = \frac{56}{13}10 = \frac{560}{13}[/itex]
[itex]P_{2A} = \frac{240}{13}2 = IV = \frac{480}{13}[/itex]
[itex]P_{10A} = IV = 10{-\frac{240}{13}-\frac{74}{13}} = \frac{3140}{13}[/itex]
These don't add up to zero. I'm kind of confused as to when you are supposed to use a negative sign or not. I used a negative sign in the last calculation because the current is going from lower potential to higher potential. Thanks for nay help you can provide.
Homework Statement
Determine if the circuit is valid or invalid.
Homework Equations
The Attempt at a Solution
Please note which node I used as the ground node and which node I have labeled [itex]V_{1}[/itex]. By the node voltage method
[itex]\frac{V_{1}}{8} + 10 - 2 + \frac{V_{1} - 10}{5} = 0[/itex]
[itex]\frac{13V_{1}}{40} + 6 = 0 [/itex]
[itex]\frac{13V_{1}}{40} = -6[/itex]
[itex]V_{1} = -\frac{240}{13} V[/itex]
[itex]I_{1} = \frac{V_{1}}{8} = -\frac{240}{13*8} = -\frac{30}{13} A[/itex]
[itex]I_{2} = \frac{V_{1}-10}{5} = \frac{\frac{-240}{13}-10}{5} = -\frac{74}{13}[/itex]
[itex]I_{3} = 10 + I_{2} = 10 + {-\frac{74}{13}} = \frac{56}{13}[/itex]
[itex]V = IR[/itex]
[itex]P = IV[/itex]
[itex]P = I^{2}R[/itex]
[itex]P_{8Ω} = I^{2}R = (-\frac{30}{13})^{2}8 = \frac{7200}{169}[/itex]
[itex]P_{5Ω} = I^{2}R = (-\frac{74}{13})^{2}5 = \frac{27380}{169}[/itex]
[itex]P_{2Ω} = I^{2}R = (\frac{56}{13})^{2}2 = \frac{6272}{169}[/itex]
[itex]P_{10V} = IV = \frac{56}{13}10 = \frac{560}{13}[/itex]
[itex]P_{2A} = \frac{240}{13}2 = IV = \frac{480}{13}[/itex]
[itex]P_{10A} = IV = 10{-\frac{240}{13}-\frac{74}{13}} = \frac{3140}{13}[/itex]
These don't add up to zero. I'm kind of confused as to when you are supposed to use a negative sign or not. I used a negative sign in the last calculation because the current is going from lower potential to higher potential. Thanks for nay help you can provide.
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