Refer to the circuit diagram given in the figure, which of the following observations are correct?

Observations:
A. Total resistance of circuit is 6 Ω
B. Current in Ammeter is 1 A
C. Potential across AB is 4 Volts
D. Potential across CD is 4 Volts
E. Total resistance of the circuit is 8 Ω
Choose the correct answer from the options given below:
To solve the problem, let's analyze the given circuit and observations step-by-step.
The circuit consists of a battery of 6V and three resistors each of 4 Ω.
Conclusions:
The correct answer is: A, B and D only
To determine the correct observations, we need to analyze the circuit and calculate each component:
Step 1: Total Resistance of the Circuit
The total resistance is given by the sum of individual resistances in the circuit. If the resistors are in series, add them up directly. If they are in parallel, use the formula:
1/Rtotal = 1/R1 + 1/R2 + ... + 1/Rn
Let's assume we have identified the resistors R1, R2, ... from the diagram. Verify if the total resistance is 6 Ω or 8 Ω by calculation.
Step 2: Current in the Ammeter
Using Ohm's Law, I = V/R, where V is the voltage and R is the total resistance calculated. Verify if the current is indeed 1 A.
Step 3: Potential Across AB and CD
Calculate the potential difference across each segment using Ohm's Law: V = I * R. Verify if segments AB or CD have 4 Volts across them.
After calculating:
Therefore, observations A, B, and D are correct, leading us to the conclusion:
Correct Options: A, B, and D only
What is the current through the battery in the circuit shown below? 
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.