The value of current $ I $ in the adjoining circuit will be 
Step 1: Analyze the circuit.
We are given a triangle circuit with three resistors, each of resistance \( 30 \, \Omega \), and a voltage source of 2 V.
Step 2: Use Ohm’s Law to find the current.
To find the current, first calculate the total equivalent resistance of the circuit. The three resistors form a delta network. First, calculate the equivalent resistance of the delta network.
Next, apply Ohm’s law \( I = \frac{V}{R} \), where \( V = 2 \, \text{V} \) and \( R_{\text{eq}} \) is the equivalent resistance. \[ R_{\text{eq}} = 90 \, \Omega \quad \text{(after simplifying the network of resistors)} \] Then, the current is: \[ I = \frac{V}{R_{\text{eq}}} = \frac{2}{90} = \frac{1}{45} \, A \]
Step 3: Conclusion.
Thus, the current \( I \) in the circuit is \( \frac{1}{45} \, A \).
Conclusion:
The correct answer is (A) \( \frac{1}{45} \, A \).
The equivalent capacitance of the circuit given between A and B is 
Let the function $ f(x) $ be defined as follows: $$ f(x) = \begin{cases} (1 + | \sin x |)^{\frac{a}{|\sin x|}}, & -\frac{\pi}{6}<x<0 \\b, & x = 0 \\ \frac{\tan 2x}{\tan 3x}, & 0<x<\frac{\pi}{6} \end{cases} $$ Then the values of $ a $ and $ b $ are: