Question 13

In the half-deflection method, let the initial current (K open) be $I_g = \frac{V}{R_1 + G}$.

When key $K$ is closed, the shunt $R_2$ is in parallel with $G$.

The current through $G$ is $I_g' = \frac{V}{R_1 + \frac{G R_2}{G + R_2}} \cdot \frac{R_2}{G + R_2} = \frac{V R_2}{R_1(G + R_2) + G R_2}$.

Given $I_g' = \frac{1}{2} I_g$, we have:

$\frac{V R_2}{R_1 G + R_1 R_2 + G R_2} = \frac{V}{2(R_1 + G)}$

$2 R_1 R_2 + 2 G R_2 = R_1 G + R_1 R_2 + G R_2$

$R_1 R_2 + G R_2 = R_1 G \Rightarrow R_1 = \frac{G R_2}{G - R_2}$

Substituting $G = 6 \Omega$ and $R_2 = 4 \Omega$:

$R_1 = \frac{6 \times 4}{6 - 4} = 12 \Omega$

In the half-deflection condition (K closed), the total resistance is $R_{eq} = R_1 + \frac{G R_2}{G + R_2} = 12 + \frac{6 \times 4}{6 + 4} = 12 + 2.4 = 14.4 \Omega$.

The current through $R_1$ is $I = \frac{V}{R_{eq}} = \frac{10}{14.4} \approx 0.69444$ A.

In mA, $I = 0.69444 \times 1000 = 694.44$ mA.