Added new codeHEADmaster

1 files changed, 29 insertions, 0 deletions

diff --git a/3830/CH1/EX1.15/Ex1_15.sce b/3830/CH1/EX1.15/Ex1_15.sce
new file mode 100644
index 000000000..febb9fc98
--- /dev/null
+++ b/3830/CH1/EX1.15/Ex1_15.sce
@@ -0,0 +1,29 @@
+// Exa 1.15

+

+clc;

+clear;

+

+// Given

+

+// Refer fig. 1.73 

+Rtotal = 11*10^6; // Total resistance in Ohms

+Vg = 1; // Voltage at gate of balancing circuit

+Vmax = 1000; // Max voltage in Volts

+// Solution

+

+printf('When the selector switch is at 1000 V position, and the mass input is 1000 V, the drop across R7 should be 1 value \n');

+R7 = (Vg/Vmax)*Rtotal;

+printf('R7 = %d k Ohms \n',R7/1000);

+printf('Similarly, when the selector switch is at the 300 V position, the drop across (R6+R7) should be 1 V \n');

+R6 = (Vg/300)*Rtotal-R7;

+printf('R6 = %.2f k Ohms \n',R6/1000);

+R5 = (Vg/100)*Rtotal-(R7+R6);

+printf('R5 = %.2f k Ohms \n',R5/1000);

+R4 = (Vg/30)*Rtotal-(R7+R6+R5);

+printf('R4 = %.2f k Ohms \n',R4/1000);

+R3 = (Vg/10)*Rtotal-(R7+R6+R5+R4);

+printf('R3 = %.2f k Ohms \n',R3/1000);

+R2 = (Vg/3)*Rtotal-(R7+R6+R5+R4+R3);

+printf('R2 = %.2f k Ohms \n',R2/1000);

+R1 = Rtotal - (R2+R3+R4+R5+R6+R7);

+printf('R1 = %.2f k Ohms \n',R1/1000);