diff options
Diffstat (limited to '1430/CH2')
30 files changed, 405 insertions, 0 deletions
diff --git a/1430/CH2/EX2.1/exa2_1.sce b/1430/CH2/EX2.1/exa2_1.sce new file mode 100644 index 000000000..0d08cd6e7 --- /dev/null +++ b/1430/CH2/EX2.1/exa2_1.sce @@ -0,0 +1,10 @@ +// Example 2.1
+// Audio Volume Control
+// It is given that amplifier draws no input current i.e i_W= 0
+// From figure 2.4, v_in(t) is related to v_s(t) by voltage divider expression
+R_AW=poly(0,'R_AW');
+p=R_AW/(5000)-0.6
+R_AW=roots(p);
+// v_in=(R_AW*v_s(t))/5000, it is required that v_out(t)=60v_s(t)
+// from above two relation we get v_in(t)=0.6v_s(t), using this relation we get R_AW
+disp(R_AW,"Value of potentiometer resistance(in Ohms)=")
diff --git a/1430/CH2/EX2.1/exa2_1.txt b/1430/CH2/EX2.1/exa2_1.txt new file mode 100644 index 000000000..f3c3878ac --- /dev/null +++ b/1430/CH2/EX2.1/exa2_1.txt @@ -0,0 +1,8 @@ +
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 2\exa2.1.sce', -1)
+
+ Value of potentiometer resistance(in Ohms)=
+
+ 3000.
+
diff --git a/1430/CH2/EX2.10/exa2_10.sce b/1430/CH2/EX2.10/exa2_10.sce new file mode 100644 index 000000000..25ed65f57 --- /dev/null +++ b/1430/CH2/EX2.10/exa2_10.sce @@ -0,0 +1,11 @@ +// Example 2.10
+// Superposition Calculations
+// First we find the contibution to i_1 from 30-V Source
+// From Figure 2.21(b)
+i_1_1=30/(6+4+2);// Ohm's Law
+// From Figure 2.22(c)
+i_1_2=(4*3)/((6+2)+4); // Current Divider
+// From Figure 2.22(d)
+i_1_3=-(6*8)/(6+(2+4)); // Current Divider
+i_1= i_1_1+i_1_2+i_1_3; // Net Current when all the Sources are active
+disp(i_1,"Net Current when all the sources are active(in Amps)=")
diff --git a/1430/CH2/EX2.10/exa2_10.txt b/1430/CH2/EX2.10/exa2_10.txt new file mode 100644 index 000000000..54a45b453 --- /dev/null +++ b/1430/CH2/EX2.10/exa2_10.txt @@ -0,0 +1,8 @@ +
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 2\exa2.10.sce', -1)
+
+ Net Current when all the sources are active(in Amps)=
+
+ - 0.5
+
diff --git a/1430/CH2/EX2.11/exa2_11.sce b/1430/CH2/EX2.11/exa2_11.sce new file mode 100644 index 000000000..6dffa28d0 --- /dev/null +++ b/1430/CH2/EX2.11/exa2_11.sce @@ -0,0 +1,10 @@ +// Example 2.11
+// Superposition with a Controlled Source
+// From figure 2.22(b)
+i_1_1=30/(6*9+(4+2));// Contribution of 30-V Source
+// From figure 2.22(c), Applying KVL in bottom Loop we get,
+// 6*(9i_1_2)+4(i_1_2-3)+2*i_1_2=0
+i_1_2=0.2;
+// When Both independent sources active,the value of i_1 is given by the sum
+i_1=i_1_1+i_1_2;
+disp(i_1,"The Value of i_1 is(in Amps)=")
diff --git a/1430/CH2/EX2.11/exa2_11.txt b/1430/CH2/EX2.11/exa2_11.txt new file mode 100644 index 000000000..6728597a8 --- /dev/null +++ b/1430/CH2/EX2.11/exa2_11.txt @@ -0,0 +1,8 @@ +
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 2\exa2.11.sce', -1)
+
+ The Value of i_1 is(in Amps)=
+
+ 0.7
+
diff --git a/1430/CH2/EX2.12/exa2_12.sce b/1430/CH2/EX2.12/exa2_12.sce new file mode 100644 index 000000000..d7b39c5f8 --- /dev/null +++ b/1430/CH2/EX2.12/exa2_12.sce @@ -0,0 +1,10 @@ +// Example 2.12
+// Thevenin Parameter from a v-i Curve
+// From v-i curve obtained from Pspice simulation we get two equations,
+x=[1 0.981;1 0.128]\[-0.491;-6.395]//Matrix Method for solving simultaneous equations
+v_oc=x(1,1);// Open-Circuit Voltage
+R_t=x(2,1);// Thevinin Resistance
+i_sc=v_oc/R_t;// Short Circuit Current
+disp(v_oc,"Open Circuit Voltage(in Volts)=")
+disp(i_sc,"Short Circuit Current(in Amps)=")
+disp(R_t,"Thevenin Resistance(in Ohms)=")
diff --git a/1430/CH2/EX2.12/exa2_12.txt b/1430/CH2/EX2.12/exa2_12.txt new file mode 100644 index 000000000..05638aee5 --- /dev/null +++ b/1430/CH2/EX2.12/exa2_12.txt @@ -0,0 +1,16 @@ +
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 2\exa2.12.sce', -1)
+
+ Open Circuit Voltage(in Volts)=
+
+ - 7.2809461
+
+ Short Circuit Current(in Amps)=
+
+ - 1.0519389
+
+ Thevenin Resistance(in Ohms)=
+
+ 6.9214537
+
diff --git a/1430/CH2/EX2.13/exa2_13.sce b/1430/CH2/EX2.13/exa2_13.sce new file mode 100644 index 000000000..3b9dc8676 --- /dev/null +++ b/1430/CH2/EX2.13/exa2_13.sce @@ -0,0 +1,17 @@ +// Example 2.13
+// Equivalent Source Networks
+//From Figure 2.28(b)
+v_oc= (20*50)/(5+20);// Open Circuit Voltage =======> equation 1
+//From Figure 2.28(c)
+i_sc=50/5; // Short Circuit Current ========> equation 2
+// From equation 1 & 2
+R_t=v_oc/i_sc;// Thevenin Resistance
+v=24;// Voltage across R_l_1
+// From figure 2.28(d)
+// Using Voltage Divider, (R_l_1*40)/(4+R_l_1)=24.
+R_l_1=6;
+//From figure 2.28(e)
+// Using Current Divider , (4*10)/(4+R_l_2)=8.
+R_l_2=1;
+disp(R_l_1,"Design Value of R_1 when v=24 (in Ohms)=")
+disp(R_l_2,"Design Value of R_1 when i=8(in Ohms)=")
diff --git a/1430/CH2/EX2.13/exa2_13.txt b/1430/CH2/EX2.13/exa2_13.txt new file mode 100644 index 000000000..3cbba4e26 --- /dev/null +++ b/1430/CH2/EX2.13/exa2_13.txt @@ -0,0 +1,12 @@ +
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 2\exa2.13.sce', -1)
+
+ Design Value of R_1 when v=24 (in Ohms)=
+
+ 6.
+
+ Design Value of R_1 when i=8(in Ohms)=
+
+ 1.
+
diff --git a/1430/CH2/EX2.14/exa2_14.sce b/1430/CH2/EX2.14/exa2_14.sce new file mode 100644 index 000000000..d11ca11be --- /dev/null +++ b/1430/CH2/EX2.14/exa2_14.sce @@ -0,0 +1,18 @@ +//Example 2.14
+// Calculating Thevenin Parameters
+// From Figure 2.31(b)
+v_x_1=0; // Applying KVL in Middle loop
+i_x_1=0;// From Ohm's Law
+i_sc=3*10^-3; // From KCL
+// For Calculating R_t
+// From figure 2.31(c)
+// v_x=-0.25v_t
+//i_x=-0.125v_t
+//i_t=i_x + v_t/40=-0.1v_s
+// R_t=v_t/i_t
+R_t=-(1*10^3)/0.1; // From equations mentioned above
+v_oc=R_t*i_sc; // Relation between thevenin Parameters
+disp(i_sc,"Short Circuit Current of Thevenin Network(in Amps)=")
+disp(R_t,"Thevenin Resistance(in Ohms)=")
+disp(v_oc,"Open Circuit Voltage of Thevenin Network(in Volts)=")
+
diff --git a/1430/CH2/EX2.14/exa2_14.txt b/1430/CH2/EX2.14/exa2_14.txt new file mode 100644 index 000000000..7b5a28606 --- /dev/null +++ b/1430/CH2/EX2.14/exa2_14.txt @@ -0,0 +1,16 @@ +
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 2\exa2.14.sce', -1)
+
+ Short Circuit Current of Thevenin Network(in Amps)=
+
+ 0.003
+
+ Thevenin Resistance(in Ohms)=
+
+ - 10000.
+
+ Open Circuit Voltage of Thevenin Network(in Volts)=
+
+ - 30.
+
diff --git a/1430/CH2/EX2.15/exa2_15.sce b/1430/CH2/EX2.15/exa2_15.sce new file mode 100644 index 000000000..e44fee2c8 --- /dev/null +++ b/1430/CH2/EX2.15/exa2_15.sce @@ -0,0 +1,11 @@ +// Example 2.15
+// Circuit Reduction by Source Conversion
+//From Figure 2.34(b)
+i_s=18; // Value of Current Source
+g_m=0.25; // Transconductance of VCCS
+G_eq=1/4+1/6+1/12;
+R_eq=1/G_eq; // Equivalent Resistance
+// Using KCL at Upper Node
+// v_2=2(18-0.25v_2)
+v_2=36/1.5;
+disp(v_2,"Voltage across 6-Ohm Resistor(in Volts)=")
diff --git a/1430/CH2/EX2.15/exa2_15.txt b/1430/CH2/EX2.15/exa2_15.txt new file mode 100644 index 000000000..1329c0ac8 --- /dev/null +++ b/1430/CH2/EX2.15/exa2_15.txt @@ -0,0 +1,8 @@ +
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 2\exa2.15.sce', -1)
+
+ Voltage across 6-Ohm Resistor(in Volts)=
+
+ 24.
+
diff --git a/1430/CH2/EX2.16/exa2_16.sce b/1430/CH2/EX2.16/exa2_16.sce new file mode 100644 index 000000000..fe0ccee1b --- /dev/null +++ b/1430/CH2/EX2.16/exa2_16.sce @@ -0,0 +1,10 @@ +// Example 2.16
+// Thevenin Network via Source Conversions
+// After Applying all Source conversion,Series & Parallel Reductions
+v_oc=-6;// Open Circuit Voltage , from figure 2.35(b)
+R_t=10;// Thevenin Resistance
+R_l=2; // Load Resistance
+i=6/(10+2); // Ohm's Law
+disp(v_oc,"Open Circuit Voltage of Thevenin Network(in Volts)=")
+disp(R_t,"Thevenin Resistance(in Ohms)=")
+disp(i,"Load Current(in Amps)=")
diff --git a/1430/CH2/EX2.16/exa2_16.txt b/1430/CH2/EX2.16/exa2_16.txt new file mode 100644 index 000000000..7176a56cc --- /dev/null +++ b/1430/CH2/EX2.16/exa2_16.txt @@ -0,0 +1,16 @@ +
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 2\exa.2.16.sce', -1)
+
+ Open Circuit Voltage of Thevenin Network(in Volts)=
+
+ - 6.
+
+ Thevenin Resistance(in Ohms)=
+
+ 10.
+
+ Load Current(in Amps)=
+
+ 0.5
+
\ No newline at end of file diff --git a/1430/CH2/EX2.2/exa2_2.sce b/1430/CH2/EX2.2/exa2_2.sce new file mode 100644 index 000000000..0ffe31320 --- /dev/null +++ b/1430/CH2/EX2.2/exa2_2.sce @@ -0,0 +1,11 @@ +// Example 2.2
+// Parallel Resistance Calculations
+// From Figure 2.6(a)
+i_x=2; // Current through unknown resistance R_x
+G_par= 1/12+ 1/24+1/8 ; // Equivalent conductance for three resistors of value 12ohms, // 24ohms & 8ohms.
+R_par=1/G_par; // Equivalent Resistance of those three resistors.
+// Using Ohm's Law and KCL
+v= R_par*(10-i_x);
+R_x= v/i_x;// Ohm's law
+disp(v,"Voltage across unknown Resistor(in Volts)=")
+disp(R_x,"Resistance across unknown Resistor(in Ohms)=")
diff --git a/1430/CH2/EX2.2/exa2_2.txt b/1430/CH2/EX2.2/exa2_2.txt new file mode 100644 index 000000000..69b4c8291 --- /dev/null +++ b/1430/CH2/EX2.2/exa2_2.txt @@ -0,0 +1,11 @@ +
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 2\exa2.2.sce', -1)
+
+ Voltage across unknown Resistor(in Volts)=
+
+ 32.
+
+ Resistance across unknown Resistor(in Ohms)=
+
+ 16.
+
diff --git a/1430/CH2/EX2.3/exa2_3.sce b/1430/CH2/EX2.3/exa2_3.sce new file mode 100644 index 000000000..6eb367cbe --- /dev/null +++ b/1430/CH2/EX2.3/exa2_3.sce @@ -0,0 +1,18 @@ +// Example 2.3
+// Available Ohmic Heating Power in Electric Grill unit
+// From figure 2.7 we know,
+R_1=12; // Resistive element 1
+R_2=24; // Resistive element 2
+v_s=120; // Voltage source
+// Switch allows one of the four resistance values given below,
+R_par= (R_1*R_2)/(R_1+R_2); // Parallel Combination
+// Individual Values of Resiators and their series combination given below
+R_ser=R_1+R_2;
+P_min=v_s^2/(R_ser); // Minimum power dissipation
+P_max=v_s^2/(R_par);// Maximum power dissipation
+// Intermediate Values of Power consumption
+P_1=v_s^2/R_1; // Power dissipated in R_1
+P_2=v_s^2/R_2; // Power dissipated in R_2
+disp(P_min,"Minimum power dissipated in Grill(in Watt)=")
+disp(P_max,"Maximum power dissipated in Grill(in Watt)=")
+disp(P_1,"&",P_2,"Intermediate values of power dissipation(in Watt)=")
diff --git a/1430/CH2/EX2.3/exa2_3.txt b/1430/CH2/EX2.3/exa2_3.txt new file mode 100644 index 000000000..4c76498ef --- /dev/null +++ b/1430/CH2/EX2.3/exa2_3.txt @@ -0,0 +1,21 @@ +
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 2\exa2.3.sce', -1)
+
+ Minimum power dissipated in Grill(in Watt)=
+
+ 400.
+
+ Maximum power dissipated in Grill(in Watt)=
+
+ 1800.
+
+ Intermediate values of power dissipation(in Watt)=
+
+ 600.
+
+ &
+
+ 1200.
+
+
diff --git a/1430/CH2/EX2.4/exa2_4.sce b/1430/CH2/EX2.4/exa2_4.sce new file mode 100644 index 000000000..39e8a342a --- /dev/null +++ b/1430/CH2/EX2.4/exa2_4.sce @@ -0,0 +1,20 @@ +// Example 2.4
+// Ladder Calculations
+//From figure 2.8(a)
+v_s=40; // Value of Voltage source
+R_par=((20*20)/(20+20))*10^3;//Equivalent resistance of two parallel 20k ohms resistors
+R_ser=(4+5+6)*10^3;//Equivalent resistance of three series resistances.
+//from figure 2.8(b)
+R_eq= 2*10^3+ ((10*15)/(10+15))*10^3;
+//The entire ladder reduces to single equivalent resistance R_eq
+// From figure 2.8(c)
+i=v_s/R_eq;// Terminal Current
+p=v_s*i;// total Power dissipated
+// from figure 2.8(b), using KVL in Left loop
+v_x=40-(2*10^3)*i; // Voltage across 20k ohm resistor
+// form figure 2.8(a),Using three-resistor voltage divider
+v_y=(5*v_x)/(4+5+6);// Voltage across 5k ohm resistor
+disp(i,"Terminal Current(in Amps)=")
+disp(p,"Total Power dissipated in Circuit(in Watt)=")
+disp(v_x,"Voltage across 20k ohm resistor(in Volts)=")
+disp(v_y,"Voltage across 5k ohm resistor(in Volts)=")
diff --git a/1430/CH2/EX2.4/exa2_4.txt b/1430/CH2/EX2.4/exa2_4.txt new file mode 100644 index 000000000..787cdb686 --- /dev/null +++ b/1430/CH2/EX2.4/exa2_4.txt @@ -0,0 +1,20 @@ +
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 2\exa2.4.sce', -1)
+
+ Terminal Current(in Amps)=
+
+ 0.005
+
+ Total Power dissipated in Circuit(in Watt)=
+
+ 0.2
+
+ Voltage across 20k ohm resistor(in Volts)=
+
+ 30.
+
+ Voltage across 5k ohm resistor(in Volts)=
+
+ 10.
+
diff --git a/1430/CH2/EX2.6/exa2_6.sce b/1430/CH2/EX2.6/exa2_6.sce new file mode 100644 index 000000000..68cc244fb --- /dev/null +++ b/1430/CH2/EX2.6/exa2_6.sce @@ -0,0 +1,15 @@ +//Example 2.6
+// Amplifier with a Field-Effect Transistor
+//g_m=5*(10^-3); // Transconductance of voltage controlled current source
+
+function[v_out]=FET(v_in)
+g_m=5*(10^-3);
+v_g=(5/6)*v_in;// from figure 2.15,using Voltage divider in Left loop
+i_out=-g_m*v_g;// Using KCL in right Loop node
+v_out=(6*10^3)*i_out;// Output voltage of Amplifier
+endfunction
+
+// For Example take v_in= 1 volt
+V_in=1;
+V_out=FET(V_in);
+disp(V_out,"Output Voltage of this amplifier(in Volts)=")
diff --git a/1430/CH2/EX2.6/exa2_6.txt b/1430/CH2/EX2.6/exa2_6.txt new file mode 100644 index 000000000..514f61cdb --- /dev/null +++ b/1430/CH2/EX2.6/exa2_6.txt @@ -0,0 +1,8 @@ +
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 2\exa2.6.sce', -1)
+
+ Output VOltage of this amplifier(in Volts)=
+
+ - 25.
+
diff --git a/1430/CH2/EX2.7/exa2_7.sce b/1430/CH2/EX2.7/exa2_7.sce new file mode 100644 index 000000000..0e73a2edd --- /dev/null +++ b/1430/CH2/EX2.7/exa2_7.sce @@ -0,0 +1,14 @@ +// Example 2.7
+// Analysis with a VCVS(Voltage controlled Voltage Source)
+// From Figure 2.16(a)
+disp("Call the function VCVS with input argument v_s")
+function[i_1]=VCVS(v_s)
+// Applying KVL on Right-hand loop we get v_2=3*i_1
+//Applying KCL at Upper node yields, i=1.5i_1
+// Using KVL in Left Loop
+i_1=v_s/(4*(1.5)+3);// v_s=4*i+v_2
+endfunction
+// For Example let us assume the value of v_s= 9 volts
+V_s=9;
+I_1=VCVS(V_s);
+disp(I_1,"Current through 10-ohm resistance(in Amps)=")
diff --git a/1430/CH2/EX2.7/exa2_7.txt b/1430/CH2/EX2.7/exa2_7.txt new file mode 100644 index 000000000..db27567bb --- /dev/null +++ b/1430/CH2/EX2.7/exa2_7.txt @@ -0,0 +1,10 @@ +
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 2\exa2.7.sce', -1)
+
+ Call the function VCVS with input argument v_s
+
+ Current through 10-ohm resistance(in Amps)=
+
+ 1.
+
diff --git a/1430/CH2/EX2.8/exa2_8.sce b/1430/CH2/EX2.8/exa2_8.sce new file mode 100644 index 000000000..244103d69 --- /dev/null +++ b/1430/CH2/EX2.8/exa2_8.sce @@ -0,0 +1,17 @@ +// Example 2.8
+// Equivalent Resistance with a VCCS(Voltage controlled Current Source)
+function[R_eq]=VCCS(R,g_m)
+// From figure 2.19, applying KCL at upper node
+// i= i_R - i_c= (v/R)-(g_m*v) = ((1-g_m*R)*v )/R
+if R*g_m == 1 then
+ R_eq= %inf; // Circuits behaves like an open circuit
+ else
+R_eq= R/(1-g_m*R);// Equivalent Resistance of the circuit
+end
+endfunction
+// For Example take g_m= 2 and R = 0.5
+funcprot(0);
+g_m= 2;
+R= 0.5;
+R_eq= VCCS(R,g_m);
+disp(R_eq,"Equivalent Resistance of the circuit(in Ohms)=")
diff --git a/1430/CH2/EX2.8/exa2_8.txt b/1430/CH2/EX2.8/exa2_8.txt new file mode 100644 index 000000000..3cad5a95c --- /dev/null +++ b/1430/CH2/EX2.8/exa2_8.txt @@ -0,0 +1,7 @@ +
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 2\exa2.8.sce', -1)
+
+ Equivalent Resistance of the circuit(in Ohms)=
+
+ Inf
diff --git a/1430/CH2/EX2.9/exa2_9.sce b/1430/CH2/EX2.9/exa2_9.sce new file mode 100644 index 000000000..64912a997 --- /dev/null +++ b/1430/CH2/EX2.9/exa2_9.sce @@ -0,0 +1,23 @@ +// Example 2.9
+// Analysis of Ladder Network Using Proportionality Principle
+// From figure 2.20
+i_1=1; // Assumption
+v_1=12*i_1;
+// Working backward toward the source using Ohm's and Kirchhoff's Laws,
+v_2=v_1/4; // Virtual Voltage across 6 ohm resistor
+i_2=v_2/6; // Virtual Current through 6 ohm resistor
+i=i_1+i_2; // Virtual Current through Independent Voltage source
+v_3=4*i;// Virtual Voltage across 3 Ohm resistor
+v_s= v_3+v_2;// Virtual Value of Independent Voltage source
+v_s_cap=72;// Actual Value of Independent Voltage source
+K=v_s_cap/v_s;
+// Actual Values of Variables are
+i_cap=K*i;
+v_2_cap=K*v_2;
+i_1_cap=K*i_1;
+R_eq= v_s_cap/i_cap; //Equivalent resistance of teh Ladder Network
+disp(i_cap,"Current through Independent Voltage Source(in Amps)=")
+disp(v_2_cap,"Voltage across 6 Ohm Resistor(in Volts)=")
+disp(i_1_cap,"Current through 12 Ohm Resistor(in Amps)=")
+disp(R_eq,"Equivalent Resistance of the Network(in Ohms)=")
+
diff --git a/1430/CH2/EX2.9/exa2_9.txt b/1430/CH2/EX2.9/exa2_9.txt new file mode 100644 index 000000000..990764c88 --- /dev/null +++ b/1430/CH2/EX2.9/exa2_9.txt @@ -0,0 +1,21 @@ +
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 2\exa2.9.sce', -1)
+
+ Current through Independent Voltage Source(in Amps)=
+
+ 12.
+
+ Voltage across 6 Ohm Resistor(in Volts)=
+
+ 24.
+
+ Current through 12 Ohm Resistor(in Amps)=
+
+ 8.
+
+ Equivalent Resistance of the Network(in Ohms)=
+
+ 6.
+
+
|