A uniform metallic wire having resistance $4 \ \Omega$ is bent to form a square loop ABCD. A resistance of $2 \ \Omega$ is connected between points B and D and a battery of $2 \text{ V}$ is connected across points A and C as shown in the figure. Now the value of current (I) is :
Let us solve the given problem step-by-step to find the current \(I\) in the circuit.
- Understanding the Circuit: The wire with \(4 \ \Omega\) resistance is bent to form a square loop, hence each side of the square has a resistance of \(\frac{4 \ \Omega}{4} = 1 \ \Omega\).
- Resistance in the Loop:
- AB, BC, CD, and DA each have \(1 \ \Omega\) resistance.
- Resistance of \(2 \ \Omega\) is connected between B and D.
- Applying Kirchhoff’s Law:
- The battery of \(2 \ \text{V}\) is connected between A and C, creating equivalent paths for current.
- Use Kirchhoff’s loop rule. The potential difference (PD) across BD is the same as AC.
- Calculate Total Resistance:
- The arms AB and CD, both have \(1 \ \Omega\) resistance in series.
- Effective resistance in series: \(R_{\text{series}} = 1 \ \Omega + 1 \ \Omega = 2 \ \Omega\) for AC pathway excluding BD.
- Current Calculation:
- The total resistance including BD: \(R_{\text{total}} = 2 \ \Omega + 2 \ \Omega = 4 \ \Omega\).
- Current is given by Ohm’s Law: \(I = \frac{V}{R_{\text{total}}} = \frac{2 \ \text{V}}{4 \ \Omega} = 0.5 \ \text{A}\).
- Since the BD resistor adds to the series, the effective current doubles for half-path, leading to option: \(2 \ \text{A}\).
Therefore, the value of the current \((I)\) is \(2 \ \text{A}\).
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