In the given circuit, the current flowing through the resistor betwee…

The correct answer is Option (1) → 0.75 A

The cell is connected between A and B, so the circuit is a set of series and parallel combinations between these two points (the cell is taken as ideal).

Readings from the figure:

    \( R_{AD} = 3\,\Omega,\ R_{DC} = 7\,\Omega,\ R_{AC} = 10\,\Omega \)
    \( R_{CB} = 5\,\Omega,\ R_{AB} = 10\,\Omega,\ V = 7.5\,\mathrm{V} \)

Path A–D–C (3 Ω + 7 Ω = 10 Ω) is in parallel with the 10 Ω diagonal AC:

    \( \displaystyle R_{AC,\,\mathrm{eq}} = \frac{10 \times 10}{10 + 10}\,\Omega = 5\,\Omega \)

This is in series with the 5 Ω resistor CB, so the branch A–C–B has 10 Ω. That branch is connected directly across the cell (in parallel with the 10 Ω resistor AB), so it has the full 7.5 V across it:

    \( \displaystyle I_{CB} = \frac{V}{R_{ACB}} = \frac{7.5\,\mathrm{V}}{10\,\Omega} = 0.75\,\mathrm{A} \)