A balanced bridge is shown in the circuit diagram. The metre bridge wire has resistance $1\ \Omega\text{m}^{-1}$. The current drawn from the battery is (Internal resistance of battery is negligible)
Choose the correct answer from the options given below
Show Hint
Always convert the wire units to meters first to match the given resistance parameter ($1\ \Omega\text{m}^{-1}$). The total meter bridge wire has a fixed structural length of exactly $1\ \text{m}$, which means its entire end-to-end baseline resistance is simple to track as a constant $1\ \Omega$ branch in parallel with the top resistors.
Step 1: Understanding the Question:
The question asks for the total current drawn from an ideal source connected across a balanced meter bridge. The wire resistance per unit length is given as $1\ \Omega\text{m}^{-1}$, and the balancing segment lengths are visible as $40\ \text{cm}$ and $60\ \text{cm}$ balancing an unknown resistor $Y$ against a $4\ \Omega$ resistor under a $6\ \text{V}$ potential supply.
Step 2: Key Formula or Approach:
1. For a balanced Wheatstone bridge network:
$$\frac{R_1}{R_2} = \frac{R_3}{R_4} \implies \frac{X}{R_{AD}} = \frac{Y}{R_{DC}}$$
2. The total resistance of a wire segment is calculated by multiplying its linear length by the resistance per unit length: $R = \text{length} \times \text{resistance rate}$.
3. Once the unknown arm resistance is found, the network simplifies to two parallel branches, allowing us to find the equivalent circuit resistance ($R_{\text{eq}}$) and apply Ohm's Law: $I = \frac{V}{R_{\text{eq}}}$.
Step 3: Detailed Explanation:
Let's first calculate the electrical resistances of the two segments of the meter wire:
Left segment length = $40\ \text{cm} = 0.4\ \text{m}$. Resistance $R_{AD} = 0.4\ \text{m} \times 1\ \Omega\text{m}^{-1} = 0.4\ \Omega$.
Right segment length = $60\ \text{cm} = 0.6\ \text{m}$. Resistance $R_{DC} = 0.6\ \text{m} \times 1\ \Omega\text{m}^{-1} = 0.6\ \Omega$.
Apply the balanced bridge balance criterion to solve for the unknown upper resistor $Y$:
$$\frac{4\ \Omega}{0.4\ \Omega} = \frac{Y}{0.6\ \Omega}$$
$$10 = \frac{Y}{0.6} \implies Y = 10 \times 0.6 = 6\ \Omega$$
Now, determine the equivalent resistance of the entire parallel network:
Top branch contains the $4\ \Omega$ and $6\ \Omega$ resistors in series: $R_{\text{top}} = 4 + 6 = 10\ \Omega$.
Bottom branch contains the two wire segments in series: $R_{\text{bottom}} = 0.4 + 0.6 = 1\ \Omega$.
Calculate the total equivalent resistance ($R_{\text{eq}}$) combining these two branches in parallel:
$$R_{\text{eq}} = \frac{R_{\text{top}} \times R_{\text{bottom}}}{R_{\text{top}} + R_{\text{bottom}}} = \frac{10 \times 1}{10 + 1} = \frac{10}{11}\ \Omega$$
Finally, apply Ohm's Law to find the total current $I$ provided by the $6\ \text{V}$ battery:
$$I = \frac{V}{R_{\text{eq}}} = \frac{6}{\frac{10}{11}} = \frac{6 \times 11}{10} = \frac{66}{10} = 0.66\ \text{A}$$
Step 4: Final Answer:
The total current drawn from the battery is 0.66 A, which corresponds to option (B).
