Given:
- Resistance of galvanometer: \( R_g = 100 \, \Omega \)
- Series resistance: \( R_s = 400 \, \Omega \)
- Voltage measured: \( V = 10 \, \text{V} \)
Step 1: Calculating the Current through the Galvanometer
The current through the galvanometer, \( i_g \), is given by:
\[ i_g = \frac{V}{R_g + R_s} = \frac{10}{400 + 100} = \frac{10}{500} = 20 \times 10^{-3} \, \text{A}. \]
Step 2: Converting the Galvanometer to an Ammeter
To convert the galvanometer into an ammeter that can read up to 10 A, we need to connect a shunt resistance \( S \) in parallel with the galvanometer. The shunt resistance is given by:
\[ i_g R_g = (I - i_g) S, \]
where \( I = 10 \, \text{A} \) is the total current.
Rearranging for \( S \):
\[ S = \frac{i_g R_g}{I - i_g} = \frac{20 \times 10^{-3} \times 100}{10 - 20 \times 10^{-3}}. \]
Simplifying:
\[ S = \frac{2}{9.98} \approx 0.2 \, \Omega. \]
Step 3: Expressing \( S \) in the Required Form
Given that \( S = x \times 10^{-2} \, \Omega \), we have:
\[ x = 20. \]
Therefore, the value of \( x \) is 20.
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Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
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