To find the resistance of a galvanometer in series or shunted circuits, use the principles of current division and Ohm's law.
Solution:
Let \( G \) be the resistance of the galvanometer. For the shunted circuit, the total current \( I \) is given by:
\[
I = 250 \, \text{mA} = 0.25 \, \text{A}.
\]
The current through the galvanometer is:
\[
I_G = \frac{I \times S}{G + S}.
\]
Substituting \( S = 5 \, \Omega \) and \( I_G = 0.25 \, \text{A} \):
\[
0.25 = \frac{0.25 \times 5}{G + 5}.
\]
Simplifying:
\[
G + 5 = 5 \quad \Rightarrow \quad G = 0 \, \Omega.
\]

The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9 V with internal resistance of 1 \(\Omega\) is connected across these points is ______ J. 
The given circuit works as: 