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1. In a common-base configuration, the input and output voltages are ________ and the output and input currents are ________.





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MCQ->In a common-base configuration, the input and output voltages are ________ and the output and input currents are ________.....
MCQ->What will be the output of the following program? #include<iostream.h> class Base { int x, y; public: Base() { x = y = 0; } Base(int xx) { x = xx; } Base(int p, int q = 10) { x = p + q; y = q; } void Display(void) { cout<< x << " " << y << endl; } }objDefault(1, 1); class Derived: public Base { Base obj; public: Derived(int xx, int yy): Base(xx, xx + 1) { } Derived(Base objB = objDefault) { } }; int main() { Derived objD(5, 3); Derived ptrD = new Derived(objD); ptrD->Display(); delete ptrD; return 0; }....
MCQ->What will be the output of the following program? #include<iostream.h> class Base { public: int S, A, M; Base(int x, int y) { S = y - y; A = x + x; M = x x; } Base(int, int y = 'A', int z = 'B') { S = y; A = y + 1 - 1; M = z - 1; } void Display(void) { cout<< S << " " << A << " " << M << endl; } }; class Derived : public Base { int x, y, z; public: Derived(int xx = 65, int yy = 66, int zz = 67): Base(x) { x = xx; y = yy; z = zz; } void Display(int n) { if(n) Base::Display(); else cout<< x << " " << y << " " << z << endl; } }; int main() { Derived objDev; objDev.Display(-1); return 0; }....
MCQ->What will be the output of the following program? #include<iostream.h> class Base { public: char S, A, M; Base(char x, char y) { S = y - y; A = x + x; M = x x; } Base(char, char y = 'A', char z = 'B') { S = y; A = y + 1 - 1; M = z - 1; } void Display(void) { cout<< S << " " << A << " " << M << endl; } }; class Derived : public Base { char x, y, z; public: Derived(char xx = 65, char yy = 66, char zz = 65): Base(x) { x = xx; y = yy; z = zz; } void Display(int n) { if(n) Base::Display(); else cout<< x << " " << y << " " << z << endl; } }; int main() { Derived objDev; objDev.Display(0-1); return 0; }....
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