A uniform wire of resistance 12 Ω 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 V and internal resistance 5 Ω 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 A
Given:
Total resistance of wire = $12 \, \Omega$
Length ratio = $1 : 2 : 3$
Resistances are proportional to length, hence:
$R_1 = 2\,\Omega,\; R_2 = 4\,\Omega,\; R_3 = 6\,\Omega$
These are connected to form a triangle. The cell (emf = 8 V, internal resistance = 5 Ω) is connected across the highest resistor $R_3 = 6\,\Omega$.
The other two resistors ($R_1 = 2\,\Omega$ and $R_2 = 4\,\Omega$) form a parallel branch across the same points as $R_3$.
Equivalent resistance of $R_1$ and $R_2$ in parallel:
$R_{12} = \frac{R_1 R_2}{R_1 + R_2} = \frac{2 \times 4}{2 + 4} = \frac{8}{6} = \frac{4}{3}\,\Omega$
Now, $R_{12}$ and $R_3$ are in parallel:
$R_{eq} = \frac{R_{12} R_3}{R_{12} + R_3} = \frac{\frac{4}{3} \times 6}{\frac{4}{3} + 6} = \frac{8}{11}\times 3 = \frac{24}{22} = 1.09\,\Omega$ (approximately)
Total resistance in circuit = $R_{eq} + r = 1.09 + 5 = 6.09\,\Omega$
Current in the circuit:
$I = \frac{E}{R_{eq} + r} = \frac{8}{6.09} = 1.31\,\text{A}$
The current through the circuit is approximately $1.3 \, \text{A}$.