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) \( \frac{1}{45} \, A \).

In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.

At 15 atm pressure, $ \text{NH}_3(g) $ is being heated in a closed container from 27°C to 347°C and as a result, it partially dissociates following the equation: $ 2\text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\text{H}_2(g) $ If the volume of the container remains constant and pressure increases to 50 atm, then calculate the percentage dissociation of $ \text{NH}_3(g) $
If equilibrium constant for the equation $ A_2 + B_2 \rightleftharpoons 2AB \quad \text{is} \, K_p, $ then find the equilibrium constant for the equation $ AB \rightleftharpoons \frac{1}{2} A_2 + \frac{1}{2} B_2. $
Consider the following reaction: $ \text{CO}(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CO}_2(g) $ At 27°C, the standard entropy change of the process becomes -0.094 kJ/mol·K. Moreover, standard free energies for the formation of $ \text{CO}_2(g) $ and $ \text{CO}(g) $ are -394.4 and -137.2 kJ/mol, respectively. Predict the nature of the above chemical reaction.