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how to practise
https://youtube.com/playlist?list=PLM-vGSGVntt1-g6SexTMe37G70Z9FmsTQ&si=2Rf4Tj1thdwJk5iU visit this channel to clear your doubt
visit this channel to clear your doubt
See lessExam ayo ksari padney ho🥲
1st sem eh..u will get used to
1st sem eh..u will get used to
See lessA question or an error I run over just now and the question is to the devs. I’m sorry this might sound rude but no downloaded files are ever in the download section.
facing same problem :(
facing same problem 🙁
See lessInstrumentation System ko question paper xa katai?
ioe syllabus app ma janus sab subejct ko latest ani oldest paper cha
ioe syllabus app ma janus sab subejct ko latest ani oldest paper cha
See lessAbout back exam
Yes milxa.. But tyo 2 tai subject same semester ko hunu parxa. For example, yedi timi le 1st sem ko 1ota ra 3rd sem ko 1ota back ko lagi apply gardai xau vane, tmilai 2000 nai pay garnu parne hunxa, but dubai subject 1st sem kai ho, ra 3rd sem kai ho vane 1000 matrai pay garnu parne hunxa
Yes milxa.. But tyo 2 tai subject same semester ko hunu parxa. For example, yedi timi le 1st sem ko 1ota ra 3rd sem ko 1ota back ko lagi apply gardai xau vane, tmilai 2000 nai pay garnu parne hunxa, but dubai subject 1st sem kai ho, ra 3rd sem kai ho vane 1000 matrai pay garnu parne hunxa
See lessElectronics Op-Amp
compare it with omp amp as adder circuit and suppose anyone one resistor eg. R3=10Kohm then its game on
compare it with omp amp as adder circuit and suppose anyone one resistor eg. R3=10Kohm then its game on
See lesshelp
insights ratni
insights ratni
See lessdiode numerical
Using KCL, a) 1 - D1 - (I1 * 1000) = 0 [For ideal diode D1 = 0] or, 1=1000 * I1 So, I1 = 0.001 A = 1mA b) 2- D2 - (I2 * 1000) = 0 [For ideal diode D2 = 0] or, 2 = 1000 * I2 So, I2 = 2mA c) 3 - D3 - (I3 * 1000) = 0 [For ideal diode D3 = 0] or, 3 = 1000 * I3 So, I3 = 3mA Total current through 1kΩ = I1Read more
Using KCL,
a) 1 – D1 – (I1 * 1000) = 0 [For ideal diode D1 = 0]
or, 1=1000 * I1
So, I1 = 0.001 A = 1mA
b) 2- D2 – (I2 * 1000) = 0 [For ideal diode D2 = 0]
or, 2 = 1000 * I2
So, I2 = 2mA
c) 3 – D3 – (I3 * 1000) = 0 [For ideal diode D3 = 0]
or, 3 = 1000 * I3
So, I3 = 3mA
Total current through 1kΩ = I1 +I2 + I3
See less= 1mA +2mA + 3mA
= 6mA
C++
#include<iostream>#include<cstdlib> #include<fstream> using namespace std; class student { public: char name[20]; int roll; float marks; public: void input( ) { cout<>name; cout<>roll; cout<>marks; } void display( ) { cout<<"\n Name= "<<name<<endRead more
#include<iostream>#include<cstdlib> #include<fstream> using namespace std; class student { public: char name[20]; int roll; float marks; public: void input( ) { cout<>name; cout<>roll; cout<>marks; } void display( ) { cout<<“\n Name= “<<name<<endl; } }; void writerecords( ) { student s[10]; fstream fout; fout.open(“student.txt”, ios :: out | ios :: app | ios :: binary); cout<<“Enter the info of 10 student for writing into binary file:\n “; for(int i = 0; i<10; i++) { s[ i ].input( ); fout.write((char *)&s[ i ], sizeof(s[ i ])) ; } fout.close( ); } void displayspecificrecord( ) { student s; fstream fin; float highest = 0; fin.open(“student.txt”, ios :: in | ios :: binary); while(fin.read((char *)&s, sizeof(s))) { if(highest < s.marks) highest=s.marks; // here we find out the highest marks among 10 students } fin.close( ); fin.open(“student.txt”, ios :: in | ios :: binary); while(fin.read((char *)&s, sizeof(s))) { if(highest==s.marks) { s.display( ); } } fin.close( ); } int main( ) { int n; while( 1 ) { cout<<“\n 1. Add record: “<<endl; cout<<“2. Display record of student whose mark is highest: “<<endl; cout<<“3. Exit: \n”; cout<>n; switch( n ) { case 1: writerecords( ); break; case 2: displayspecificrecord( ); break; case 3: exit(0); default: cout<<“The choice you entered is wrong: “; } } return 0; }
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