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 figure shows an opamp circuit with a 5.1 V Zener diode in the feedback loop. The opamp runs from \( \pm 15 \, {V} \) supplies. If a \( +1 \, {V} \) signal is applied at the input, the output voltage (rounded off to one decimal place) is:
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.
State Kirchhoff's law related to electrical circuits. In the given metre bridge, balance point is obtained at D. On connecting a resistance of 12 ohm parallel to S, balance point shifts to D'. Find the values of resistances R and S.