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<P><p> </p></P></DIV></TD></TR></TABLE></v:textbox></v:shape><w:wrap type="topAndBottom"></w:wrap></v:group></wrapblock><BR clear=all >電路圖的方程式為:
<P> -V<SUB>1</SUB> + R<SUB>1</SUB>i<SUB>1 </SUB>+ R<SUB>4 </SUB>= 0</P>
<P> -R<SUB>4</SUB>i<SUB>4 </SUB>+ R<SUB>2</SUB>i<SUB>2 </SUB>+<SUB> </SUB>R<SUB>5</SUB>i<SUB>5 </SUB>= 0</P>
<P> -R<SUB>5</SUB>i<SUB>5 </SUB>+ R<SUB>3</SUB>i<SUB>3 </SUB>+ V<SUB>2 </SUB>= 0</P>
<P > i<SUB>1 </SUB>= i<SUB>2 </SUB>+ i<SUB>4</SUB> </P>
<P > i<SUB>2 </SUB>= i<SUB>3 </SUB>+ i<SUB>5<p></p></SUB></P>
<P >. </P>
<P >a. 給定的電阻值R<SUB>1 </SUB>= 5KΩ, R<SUB>2 </SUB>= 100KΩ, R<SUB>3 </SUB>= 200KΩ, R<SUB>4 </SUB>= 150KΩ, R<SUB>5 </SUB>= 250KΩ(注意: 1KΩ=10001KΩ) ,當電壓值V<SUB>1</SUB>=100, V<SUB>2</SUB>=50,求出i<SUB>1 </SUB>,i<SUB>2</SUB>,i<SUB>3</SUB>的值<p></p></P>
<P >b. 給定的電阻值R<SUB>1 </SUB>= 5KΩ, R<SUB>2 </SUB>= 100KΩ, R<SUB>3 </SUB>= 200KΩ, R<SUB>4 </SUB>= 150KΩ, R<SUB>5 </SUB>= 250KΩ,電壓V<SUB>1</SUB>=100伏特.假設每個電阻額定的承載量不能大於1毫安培(=0.001安培),計算出電壓V<SUB>2</SUB>的允許範圍.<p></p></P>
<P >c. 假設我們想要知道電阻R<SUB>3</SUB>限制V<SUB>2</SUB>允許範圍的情況,求岀V<SUB>2</SUB>允許範圍為R<SUB>3</SUB>的函數, 150 < R<SUB>3 </SUB>< 250 KΩ,並加以繪圖.<p></p></P>
<P>我利用了i<SUB>1 </SUB>= i<SUB>2 </SUB>+ i<SUB>4</SUB> 及i<SUB>2 </SUB>= i<SUB>3 </SUB>+ i<SUB>5</SUB>這兩個式子將前面三個方程式中的i<SUB>4</SUB>與i<SUB>5</SUB>消掉得到了三個新的方程式 </P>
<P >問題a我利用了i<SUB>1 </SUB>= i<SUB>2 </SUB>+ i<SUB>4</SUB> 及i<SUB>2 </SUB>= i<SUB>3 </SUB>+ i<SUB>5</SUB>這兩個式子將前面三個方程式中的i<SUB>4</SUB>與i<SUB>5</SUB>消掉得到了三個新的方程式<p></p></P>
<P> (R<SUB>1</SUB>+ R<SUB>4</SUB>)i<SUB>1</SUB> - R<SUB>4</SUB>i<SUB>2 </SUB>= V<SUB>1</SUB><p></p></P>
<P >- R<SUB>4</SUB>i<SUB>1 </SUB>+ (R<SUB>2</SUB>+ R<SUB>4</SUB>+ R<SUB>5</SUB>)i<SUB>2 </SUB>-<SUB> </SUB>R<SUB>5</SUB>i<SUB>3 </SUB>= 0<p></p></P>
<P >R<SUB>5</SUB>i<SUB>2 </SUB>- (R<SUB>3</SUB>+ R<SUB>5</SUB>)i<SUB>3 </SUB>= V<SUB>2</SUB><p></p></P>
<P> 寫出程式解i<SUB>1 </SUB>,i<SUB>2</SUB>,i<SUB>3</SUB>的值<p></p></P>
<P>>>R=[5,100,200,150,250]*1000;<p></p></P>
<P>>>V1=100;V2=50;<p></p></P>
<P>>>A1=[R(1)+R(4),-R(4),0];<p></p></P>
<P>>>A2=[-R(4),R(2)+R(4)+R(5),-R(5)];<p></p></P>
<P>>>A3=[0,R(5),-(R(3)+R(5))];<p></p></P>
<P>>>A=[A1;A2;A3];<p></p></P>
<P>>>b=[V1;0;V2];<p></p></P>
<P>>>current=A/b;<p></p></P>
<P>>>disp(‘The currents are:’)<p></p></P>
<P>>>disp(current)<p></p></P>
<P>>>resist<p></p></P>
<P>The current are:<p></p></P>
<P >1.0e-0.03*<p></p></P>
<P> 0.9544<p></p></P>
<P> 0.3195<p></p></P>
<P> 0.0664<p></p></P>
<P>所得解i<SUB>1 </SUB>=0.9544mA,i<SUB>2</SUB>=0.3195mA,i<SUB>3</SUB>=0.0664mA<p></p></P>
<P>然而問題b,c由於相關到回圈問題由於初學的關係我寫不太出來作出的答案好像都怪怪的,所以想求救於各位,請求協助幫忙...謝謝</P> |