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?Applying Kirchhoff's Laws to the Electrical Network in the Figure,the

Question 47

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?Applying Kirchhoff's Laws to the electrical network in the figure,the currents I1,I2,and I3 are the solution of the system {I1I2+I3=06I1+3I2=113I2+5I3=6\left\{ \begin{array} { l l } I _ { 1 } - I _ { 2 } + I _ { 3 } & = 0 \\6 I _ { 1 } + 3 I _ { 2 } & = 11 \\3 I _ { 2 } + 5 I _ { 3 } & = 6\end{array} \right. Find the currents.?  ?Applying Kirchhoff's Laws to the electrical network in the figure,the currents I<sub>1</sub>,I<sub>2</sub>,and I<sub>3</sub> are the solution of the system  \left\{ \begin{array} { l l }  I _ { 1 } - I _ { 2 } + I _ { 3 } & = 0 \\ 6 I _ { 1 } + 3 I _ { 2 } & = 11 \\ 3 I _ { 2 } + 5 I _ { 3 } & = 6 \end{array} \right.  Find the currents.?    \begin{array} { l }  V _ { 1 } = 11 \text { volts, } V _ { 2 } = 6 \text { volts } \\ R _ { 1 } = 6 \Omega , R _ { 2 } = 3 \Omega , R _ { 3 } = 5 \Omega \end{array}  ? A) ?I<sub>1</sub> =  - \frac { 40 } { 3 }  <sub> </sub>amperes;I<sub>2</sub><sub> </sub>=  \frac { 91 } { 3 }  amperes;I<sub>3</sub> = -17 amperes B) ?I<sub>1</sub> =  \frac { 10 } { 9 }  amperes;I<sub>2</sub> =  \frac { 13 } { 9 }  amperes;I<sub>3</sub> =  \frac { 1 } { 3 }  amperes C) ?I<sub>1</sub> = 6 amperes;I<sub>2</sub> = 3 amperes;I<sub>3</sub> = 5 amperes D) ?I<sub>1</sub> =  \frac { 70 } { 33 }  amperes;I<sub>2</sub> =  - \frac { 19 } { 33 }  amperes;I<sub>3</sub> =  \frac { 17 } { 11 }  amperes E) ?I<sub>1</sub> = 70 amperes;I<sub>2</sub> = 91 amperes;I<sub>3</sub> = 21 amperes V1=11 volts, V2=6 volts R1=6Ω,R2=3Ω,R3=5Ω\begin{array} { l } V _ { 1 } = 11 \text { volts, } V _ { 2 } = 6 \text { volts } \\R _ { 1 } = 6 \Omega , R _ { 2 } = 3 \Omega , R _ { 3 } = 5 \Omega\end{array} ?


A) ?I1 = 403- \frac { 40 } { 3 } amperes;I2 = 913\frac { 91 } { 3 } amperes;I3 = -17 amperes
B) ?I1 = 109\frac { 10 } { 9 } amperes;I2 = 139\frac { 13 } { 9 } amperes;I3 = 13\frac { 1 } { 3 } amperes
C) ?I1 = 6 amperes;I2 = 3 amperes;I3 = 5 amperes
D) ?I1 = 7033\frac { 70 } { 33 } amperes;I2 = 1933- \frac { 19 } { 33 } amperes;I3 = 1711\frac { 17 } { 11 } amperes
E) ?I1 = 70 amperes;I2 = 91 amperes;I3 = 21 amperes

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