Q9 (2024)
A uniform metallic wire having resistance $4\Omega$ is bent to form a square loop (ABCD) (see figure). A resistance of $2\Omega$ is connected between points B and D and a battery of 2 V is connected across points A and C as shown in the figure. Now the value of current (I) is
{{IMG1}}
- 2 A
- 1 A
- 3 A
- 4.5 A
### Related Formula
$$I = \frac{E}{R_{\text{eff}}}$$
### Core Logic
Each side of the square loop has resistance $1\Omega$.
{{SOL_IMG1}}
This constitutes a balanced Wheatstone bridge between points A and C.
No current flows through the central $2\Omega$ resistor.
{{SOL_IMG2}}
$$R_{\text{eff, AC}} = \frac{(1+1) \times (1+1)}{(1+1) + (1+1)} = 1\Omega$$
$$I = \frac{E}{R_{\text{eff}}} = \frac{2}{1} = 2\mathrm{A}$$
### Pattern Recognition
Symmetrical bridge configuration leads to zero current in central resistor.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
