The diagram below illustrates a metre bridge with an unknown resistance X and 30 in the arms respectively. If the balance point is 40 cm from A, calculate the magnitude of X.

In a balanced metre bridge, the ratio of resistances is equal to the ratio of the lengths of the wire segments.
Given:

• Resistance $R_1 = 30 \, \Omega$
• Length $L_1 = 40 \, \text{cm}$ (from A to the balance point)
• Total length of the wire is $100 \, \text{cm}$.
• Therefore, the other length $L_2 = 100 \, \text{cm} - L_1 = 100 \, \text{cm} - 40 \, \text{cm} = 60 \, \text{cm}$.
• Unknown resistance $R_2 = X$.

The formula for a balanced metre bridge is:

$$\frac{R_1}{R_2} = \frac{L_1}{L_2}$$

$$\frac{30}{X} = \frac{40}{60}$$ $$\frac{30}{X} = \frac{2}{3}$$ $$2X = 30 \times 3$$ $$2X = 90$$ $$X = \frac{90}{2}$$ $$X = 45 \, \Omega$$

The magnitude of X is $\boxed{\text{45 } \Omega}$.

Here's the phasor diagram for the alternating currents:

(i) Two alternating currents $I_1$ and $I_2$ are defined by the expression $I_1 = I_0 \sin(\omega t)$ and $I_2 = I_0 \cos(\omega t)$ respectively. Draw the phasor diagram to represent the currents $I_1$ and $I_2$.

We can express $I_1$ in terms of cosine: $I_1 = I_0 \sin(\omega t) = I_0 \cos(\omega t - \frac{\pi}{2})$.
This means $I_1$ lags $I_2$ by $\frac{\pi}{2}$ (or $90^\circ$).

The phasor diagram will show $I_2$ along the positive x-axis (representing $\cos(\omega t)$) and $I_1$ along the negative y-axis (representing $\cos(\omega t - \frac{\pi}{2})$). Both phasors have the same amplitude $I_0$.

\begin{tikzpicture} % Axes \draw[->] (-2,0) -- (2,0) node[right] {Real Axis}; \draw[->] (0,-2) -- (0,2) node[above] {Imaginary Axis}; % Phasor I2 \draw[->, thick, blue] (0,0) -- (1.5,0) node[above right] {$I_2 = I_0 \cos(\omega t)$}; % Phasor I1 \draw[->, thick, red] (0,0) -- (0,-1.5) node[below right] {$I_1 = I_0 \sin(\omega t)$}; % Angle \draw[dashed] (0,0) -- (1.5,0); \draw[dashed] (0,0) -- (0,-1.5); \draw[<->, gray] (0.5,0) arc (0:-90:0.5); \node[gray] at (0.7,-0.7) {$90^\circ$}; \end{tikzpicture}