For the network shown in the given figure, if no current flows through the branch AO, the current drawn from the cell, is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2A
The resistors 4 Ω, R, 2 Ω and 2.5 Ω form a Wheatstone bridge, with the branch AO as the galvanometer arm. No current flows in AO, so the bridge is balanced: the ratio of the two resistors on one side equals the ratio on the other.
\( \displaystyle \frac{4}{R} = \frac{2}{2.5} \Rightarrow R = \frac{4\times2.5}{2} = 5\,\Omega \)
The branch AO carries no current and can be removed. The cell (10 V, with a 2 Ω resistor in series with it) then feeds two parallel paths:
\( R_{1} = 4+5 = 9\,\Omega,\quad R_{2} = 2+2.5 = 4.5\,\Omega \)
\( \displaystyle R_{\mathrm{par}} = \frac{9\times4.5}{9+4.5}\,\Omega = 3\,\Omega \)
\( R_{\mathrm{total}} = 3+2 = 5\,\Omega \)
\( \displaystyle I = \frac{V}{R_{\mathrm{total}}} = \frac{10\,\mathrm{V}}{5\,\Omega} = 2\,\mathrm{A} \)