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Consider the Circuit Shown in the Figure Below I1,I2, and I3I _ { 1 } , I _ { 2 } , \text { and } I _ { 3 }

Question 74

Multiple Choice

Consider the circuit shown in the figure below. The currents I1,I2, and I3I _ { 1 } , I _ { 2 } , \text { and } I _ { 3 } ( in amperes ) are the solutions to the system of linear equations {3I1+6I3=E12I2+6I3=E2I1+I2I3=0\left\{ \begin{aligned}3 I _ { 1 } + 6 I _ { 3 } & = E _ { 1 } \\2 I _ { 2 } + 6 I _ { 3 } & = E _ { 2 } \\I _ { 1 } + I _ { 2 } - I _ { 3 } & = 0\end{aligned} \right. , where E1 and E2E _ { 1 } \text { and } E _ { 2 } are voltages of 6 and 9 volts, respectively. Use the inverse of the coefficient matrix of this system to find the unknown currents. Round answers to nearest tenth. [Hint: If a current value turns out to be negative, it simply means that the current flow is in the opposite direction from that indicated in the figure below.]  Consider the circuit shown in the figure below. The currents  I _ { 1 } , I _ { 2 } , \text { and } I _ { 3 }  ( in amperes )  are the solutions to the system of linear equations  \left\{ \begin{aligned} 3 I _ { 1 } + 6 I _ { 3 } & = E _ { 1 } \\ 2 I _ { 2 } + 6 I _ { 3 } & = E _ { 2 } \\ I _ { 1 } + I _ { 2 } - I _ { 3 } & = 0 \end{aligned} \right.  , where  E _ { 1 } \text { and } E _ { 2 }  are voltages of 6 and 9 volts, respectively. Use the inverse of the coefficient matrix of this system to find the unknown currents. Round answers to nearest tenth. [Hint: If a current value turns out to be negative, it simply means that the current flow is in the opposite direction from that indicated in the figure below.]     A)   I _ { 1 } = 0.5 \mathrm { amps } ; I _ { 2 } = 0.8 \mathrm { amps } ; I _ { 3 } = 1.3 \mathrm { amps }  B)   I _ { 1 } = - 0.4 \mathrm { amps } ; I _ { 2 } = 1.6 \mathrm { amps } ; I _ { 3 } = 1.2 \mathrm { amps }  C)   I _ { 1 } = 1.3 \mathrm { amps } ; I _ { 2 } = 6.4 \mathrm { amps } ; I _ { 3 } = - 1.3 \mathrm { amps }  D)   I _ { 1 } = - 0.2 \mathrm { amps } ; I _ { 2 } = 1.3 \mathrm { amps } ; I _ { 3 } = 1.1 \mathrm { amps }  E)   I _ { 1 } = - 4.9 \mathrm { amps } ; I _ { 2 } = 15.0 \mathrm { amps } ; I _ { 3 } = 8.2 \mathrm { amps }  Consider the circuit shown in the figure below. The currents  I _ { 1 } , I _ { 2 } , \text { and } I _ { 3 }  ( in amperes )  are the solutions to the system of linear equations  \left\{ \begin{aligned} 3 I _ { 1 } + 6 I _ { 3 } & = E _ { 1 } \\ 2 I _ { 2 } + 6 I _ { 3 } & = E _ { 2 } \\ I _ { 1 } + I _ { 2 } - I _ { 3 } & = 0 \end{aligned} \right.  , where  E _ { 1 } \text { and } E _ { 2 }  are voltages of 6 and 9 volts, respectively. Use the inverse of the coefficient matrix of this system to find the unknown currents. Round answers to nearest tenth. [Hint: If a current value turns out to be negative, it simply means that the current flow is in the opposite direction from that indicated in the figure below.]     A)   I _ { 1 } = 0.5 \mathrm { amps } ; I _ { 2 } = 0.8 \mathrm { amps } ; I _ { 3 } = 1.3 \mathrm { amps }  B)   I _ { 1 } = - 0.4 \mathrm { amps } ; I _ { 2 } = 1.6 \mathrm { amps } ; I _ { 3 } = 1.2 \mathrm { amps }  C)   I _ { 1 } = 1.3 \mathrm { amps } ; I _ { 2 } = 6.4 \mathrm { amps } ; I _ { 3 } = - 1.3 \mathrm { amps }  D)   I _ { 1 } = - 0.2 \mathrm { amps } ; I _ { 2 } = 1.3 \mathrm { amps } ; I _ { 3 } = 1.1 \mathrm { amps }  E)   I _ { 1 } = - 4.9 \mathrm { amps } ; I _ { 2 } = 15.0 \mathrm { amps } ; I _ { 3 } = 8.2 \mathrm { amps }


A) I1=0.5amps;I2=0.8amps;I3=1.3ampsI _ { 1 } = 0.5 \mathrm { amps } ; I _ { 2 } = 0.8 \mathrm { amps } ; I _ { 3 } = 1.3 \mathrm { amps }
B) I1=0.4amps;I2=1.6amps;I3=1.2ampsI _ { 1 } = - 0.4 \mathrm { amps } ; I _ { 2 } = 1.6 \mathrm { amps } ; I _ { 3 } = 1.2 \mathrm { amps }
C) I1=1.3amps;I2=6.4amps;I3=1.3ampsI _ { 1 } = 1.3 \mathrm { amps } ; I _ { 2 } = 6.4 \mathrm { amps } ; I _ { 3 } = - 1.3 \mathrm { amps }
D) I1=0.2amps;I2=1.3amps;I3=1.1ampsI _ { 1 } = - 0.2 \mathrm { amps } ; I _ { 2 } = 1.3 \mathrm { amps } ; I _ { 3 } = 1.1 \mathrm { amps }
E) I1=4.9amps;I2=15.0amps;I3=8.2ampsI _ { 1 } = - 4.9 \mathrm { amps } ; I _ { 2 } = 15.0 \mathrm { amps } ; I _ { 3 } = 8.2 \mathrm { amps }

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