To solve the problem, we need to find the resistance to be connected in series with the tangent galvanometer in order to get a deflection of \(30^{\circ}\). The tangent law of a galvanometer states that the tangent of the deflection angle (\(\theta\)) is proportional to the current (\(I\)) flowing through it:
\(I = k \cdot \tan(\theta)\),
where \(k\) is a constant of proportionality.
Initially, we have:
The total resistance in the initial condition is the sum of the internal resistance of the battery and the resistance of the galvanometer:
\(R_{\text{total1}} = r + R_g = 2 + 4 = 6\Omega\)
The current through the galvanometer for deflection \(60^{\circ}\) is:
\(I_1 = \frac{E}{R_{\text{total1}}} = \frac{12}{6} = 2 \text{A}\)
Applying tangent law:
\(I_1 = k \cdot \tan(60^{\circ}) = k \cdot \sqrt{3}\)
Thus, \(k = \frac{2}{\sqrt{3}}\)
For the second condition where deflection is \(30^{\circ}\):
\(I_2 = k \cdot \tan(30^{\circ}) = \frac{2}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} = \frac{2}{3} \text{A}\)
To find the required total resistance to have a deflection of \(30^{\circ}\), we use:
\(R_{\text{total2}} = \frac{E}{I_2} = \frac{12}{\frac{2}{3}} = 18\Omega\)
The additional resistance \(R\) required in series is given by:
\(R = R_{\text{total2}} - R_{\text{total1}} = 18 - 6 = 12\Omega\)
Therefore, the resistance to be connected in series with the tangent galvanometer to obtain a deflection of \(30^{\circ}\) is 12\(\Omega\).