
In the given circuit, we have two diodes \( D_1 \) and \( D_2 \). The voltage across the diodes and the resistor is affected by the direction in which the diodes are connected and their characteristics.
- \( D_1 \) is forward-biased because its anode is connected to +5V, and the cathode is connected to the node where \( V_o \) is measured.
- \( D_2 \) is reverse-biased because its anode is connected to the node \( V_o \), and its cathode is connected to ground.
Since \( D_2 \) is reverse-biased, it will not conduct, and \( D_1 \) will conduct. Therefore, the output voltage \( V_o \) will be zero because the voltage drop across the conducting diode \( D_1 \) is almost zero in a forward-biased condition.
Thus, the output voltage is zero.
Therefore, the correct answer is (2) Zero.
For a given reaction \( R \rightarrow P \), \( t_{1/2} \) is related to \([A_0]\) as given in the table. Given: \( \log 2 = 0.30 \). Which of the following is true?
| \([A]\) (mol/L) | \(t_{1/2}\) (min) |
|---|---|
| 0.100 | 200 |
| 0.025 | 100 |
A. The order of the reaction is \( \frac{1}{2} \).
B. If \( [A_0] \) is 1 M, then \( t_{1/2} \) is \( 200/\sqrt{10} \) min.
C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.
D. \( t_{1/2} \) is 800 min for \( [A_0] = 1.6 \) M.