A uniform wire of resistance \( 12\,\Omega \) is cut into three pieces in the ratio of length 1: 2: 3. Now the three pieces are connected to form a triangle. A cell of emf \( 8\,\mathrm{V} \) and internal resistance \( 5\,\Omega \) is connected across the highest of the three resistors. The current through the circuit is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 1\,\mathrm{A} \)
The resistance of a uniform wire is proportional to its length, so the three pieces have resistances in the ratio 1 : 2 : 3, and their total is \( 12\,\Omega \).
\( \displaystyle R_1 = 2\,\Omega,\quad R_2 = 4\,\Omega,\quad R_3 = 6\,\Omega \)
The pieces form a triangle. The cell is connected across the highest resistor, \( 6\,\Omega \). The other two pieces are in series with each other, and this pair is in parallel with the \( 6\,\Omega \) piece.
\( \displaystyle R_{\mathrm{ext}} = \frac{6 \times (2 + 4)}{6 + (2 + 4)} = \frac{36}{12} = 3\,\Omega \)
The internal resistance is in series with this combination.
\( \displaystyle I = \frac{\varepsilon}{R_{\mathrm{ext}} + r} = \frac{8}{3 + 5} = 1\,\mathrm{A} \)