The resistance of each arm of the wheat stone bridge is `10Omega`. A resistance of `10Omega` is connected in series with galvanometer then the equivalent resistance across the battery will be:-

Options

A. 15
B. 25
C. 45
D. 55

Explanation

To solve the problem, we need to analyze the Wheatstone bridge configuration and the additional resistance connected in series with the galvanometer. Here’s a step-by-step solution:

### Step 1: Understand the Wheatstone Bridge Configuration
A Wheatstone bridge consists of four resistances arranged in a diamond shape. In this case, each arm of the bridge has a resistance of \(10 \, \Omega\).

### Step 2: Identify the Balanced Condition
For a balanced Wheatstone bridge, the ratio of the resistances in one pair of opposite arms is equal to the ratio in the other pair. Since all arms have the same resistance of \(10 \, \Omega\), the bridge is balanced. This means the current through the galvanometer is zero.

### Step 3: Simplify the Circuit
Since the galvanometer does not carry any current (due to the balanced condition), we can ignore the galvanometer and focus on the resistances in the bridge. The two pairs of resistances can be combined:
- The two \(10 \, \Omega\) resistors on one side of the bridge are in series, giving a total of \(10 + 10 = 20 \, \Omega\).
- The same applies to the other side, resulting in another \(20 \, \Omega\).

### Step 4: Combine the Series Resistances
Now, we have two resistances of \(20 \, \Omega\) each, which are in parallel with each other. The equivalent resistance \(R_{eq}\) for two resistors in parallel is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
Substituting \(R_1 = 20 \, \Omega\) and \(R_2 = 20 \, \Omega\):
\[ \frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \]
Thus,
\[ R_{eq} = 10 \, \Omega \]

### Step 5: Add the Series Resistance
Now, we need to account for the additional \(10 \, \Omega\) resistance that is connected in series with the galvanometer. The total equivalent resistance \(R_{total}\) across the battery will be:
\[ R_{total} = R_{eq} + R_{series} = 10 \, \Omega + 10 \, \Omega = 20 \, \Omega \]

### Final Answer
The equivalent resistance across the battery is \(20 \, \Omega\).

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