Problem set 78

1. The experimentally measured spin $g$ factors of a proton and a neutron indicate that
1. both proton and neutron are elementary point particles
2. both proton and neutron are not elementary point particles
3. while proton is an elementary point particle, neutron is not
4. while neutron is an elementary point pat1icle, proton is not

both proton and neutron are not elementary point particles

Hence, answer is (B)

2. The tank circuit of a Hartley oscillator is shown in the figure. If $M$ is the mutual inductance between the inductors, the oscillation frequency is
   
1. $\frac{1}{2\pi\sqrt{(L_1+L_2+2M)C}}$
2. $\frac{1}{2\pi\sqrt{(L_1+L_2-2M)C}}$
3. $\frac{1}{2\pi\sqrt{(L_1+L_2+M)C}}$
4. $\frac{1}{2\pi\sqrt{(L_1+L_2-M)C}}$

Frequency of Hartley oscillator is given by $$f=\frac{1}{2\pi\sqrt{LC}}$$ where $L=L_1+L_2$ if the coils are assumed to be winded on different cores and $L_1+L_2+2M$ if they are winded on a single core.

Hence, answer is (A)

3. In the given digital logic circuit, $A$ and $B$ form the input. The output $Y$ is
   
1. $Y=\bar A$
2. $Y=A\bar B$
3. $Y=A\oplus B$
4. $Y=\bar B$

\begin{align*} Y&=\overline{\left(\bar A+B\right) \left( A+B\right)}\\ &=\overline{\left(\bar A+B\right)}+ \overline{\left( A+B\right)}\\ &=\overline{\left(\bar A\right)}\bar B+\bar A\bar B\\ &=A\bar B+\bar A\bar B\\ &=\left(A+\bar A\right)\bar B\\ &=\bar B\quad \text{since, }(A+\bar A=1) \end{align*}

Hence, answer is (D)

4. The largest analog output voltage from a 6-bit digital to analog converter (DAC) which produces 1.0 V output for a digital input of 010100, is
1. 1.6 V
2. 2.9 V
3. 3.15 V
4. 5.0 V

For a n-bit DAC $$V_{out}={\textstyle V_R\left(\frac{b_{n-1}}{2^1}+\frac{b_{n-2}}{2^2}+\cdots+\frac{b_{0}}{2^n}\right)}$$ For a 6-bit DAC $$V_{out}=\!\!{\textstyle V_R\!\!\left(\frac{b_{5}}{2^1}\!+\!\frac{b_{4}}{2^2}\!+\!\frac{b_{3}}{2^3}\!+\!\frac{b_{2}}{2^4}\!+\!\frac{b_{1}}{2^5}\!+\!\frac{b_{0}}{2^6}\right)}$$ For 1.0 V output of a digital input of 010100 $$1=\!{\textstyle V_R\!\!\left(\frac{0}{2^1}+\frac{1}{2^2}+\frac{0}{2^3}+\frac{1}{2^4}+\frac{0}{2^5}+\frac{0}{2^6}\right)}$$ $$V_R=\frac{16}{5}$$ $$V_{max}\!=\!{\textstyle \frac{16}{5}\!\left(\frac{1}{2^1}\!+\!\frac{1}{2^2}\!+\!\frac{1}{2^3}\!+\!\frac{1}{2^4}+\frac{1}{2^5}\!+\!\frac{1}{2^6}\right)}$$ $$V_{max}=3.15 V$$

Hence, answer is (C)

5. The low-pass active filter shown in the figure has a cut-off frequency of 2 kHz and a pass band gain of 1.5. The values of the resistors are
   
1. $R_1 = 10\: k\Omega$; $R_2 = 1.3 \Omega$
2. $R_1 = 30\: k\Omega$; $R_2 = 1.3 \Omega$
3. $R_1 = 10\: k\Omega$; $R_2 = 1.7 k\Omega$
4. $R_1 = 30\: k\Omega$; $R_2 = 1.7 k\Omega$

Using $$Gain=1+\frac{15k}{R_1}$$ $$R_1=30\:k\Omega$$ Using $$f_c=\frac{1}{2\pi RC}$$ $$R=\frac{1}{2\pi f_cC}=1.7\:k\Omega$$

Hence, answer is (D)