Current Electricity — Medium JEE physics MCQ
In the circuit shown, each resistor has resistance $R$. The equivalent resistance between points A and B is:
```
A ---[R]---+---[R]---B
|
[R]
|
-------+-------
```
- A. $\frac{R}{3}$
- B. $\frac{2R}{3}$
- C. $\frac{3R}{2}$
- D. $2R$
Solution
The middle resistor $R$ is in series with the parallel combination of two resistors.
Wait, let me reconsider the circuit. We have two resistors in series (top path) giving $2R$, and this is in parallel with one resistor $R$.
$R_{eq} = \frac{2R \times R}{2R + R} = \frac{2R^2}{3R} = \frac{2R}{3}$
Actually, looking at the circuit again: two $R$ in the top branch (series = $2R$), parallel with $R$ in middle:
$\frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{R} = \frac{1 + 2}{2R} = \frac{3}{2R}$
$R_{eq} = \frac{2R}{3}$
The correct option is **B**.