From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001 From: prashantsinalkar Date: Tue, 10 Oct 2017 12:27:19 +0530 Subject: initial commit / add all books --- 2459/CH5/EX5.5/Ex5_5.PNG | Bin 0 -> 6094 bytes 2459/CH5/EX5.5/Ex5_5.sce | 21 +++++++++++++++++++++ 2459/CH5/EX5.5/Figure5_5.JPG | Bin 0 -> 21276 bytes 3 files changed, 21 insertions(+) create mode 100644 2459/CH5/EX5.5/Ex5_5.PNG create mode 100644 2459/CH5/EX5.5/Ex5_5.sce create mode 100644 2459/CH5/EX5.5/Figure5_5.JPG (limited to '2459/CH5/EX5.5') diff --git a/2459/CH5/EX5.5/Ex5_5.PNG b/2459/CH5/EX5.5/Ex5_5.PNG new file mode 100644 index 000000000..bdd4596fe Binary files /dev/null and b/2459/CH5/EX5.5/Ex5_5.PNG differ diff --git a/2459/CH5/EX5.5/Ex5_5.sce b/2459/CH5/EX5.5/Ex5_5.sce new file mode 100644 index 000000000..dcc87b75f --- /dev/null +++ b/2459/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,21 @@ +//chapter5 +//example5.5 +//page95 + +rp=1000 // ohm +Rl=10 // ohm +Eg=8 // V +mu=20 + +// the diagram in book is for understanding only. Also we do not have a block of "triode" in scilab xcos. The figure is not required to solve the problem. +// however, the equivalent circuit has been drawn in xcos for reference. + +// since rp=n^2*Rl for maximum power transfer so +n=(rp/Rl)^0.5 + +// P_max=Ip^2*RE where Ip=mu*Eg/(rp+RE) and RE=rp +// thus +P_max=((mu*Eg)^2)/(4*rp) + +printf("transformation ratio n= %.2f \n",n) +printf("power supplied to speaker when signal is 8V rms is = %.3f W",P_max) diff --git a/2459/CH5/EX5.5/Figure5_5.JPG b/2459/CH5/EX5.5/Figure5_5.JPG new file mode 100644 index 000000000..172a31849 Binary files /dev/null and b/2459/CH5/EX5.5/Figure5_5.JPG differ -- cgit