The terminal voltage and current of a linear electrical network shown in Figure (a) are given in the table.

$$ \begin{array}{|c|c|} \hline \text { Terminal voltage }\left(v_t\right) & \text { Terminal current }\left(i_t\right) \\ \hline 18 \mathrm{~V} & -0.5 \mathrm{~A} \\ \hline 30 \mathrm{~V} & 0.5 \mathrm{~A} \\ \hline 36 \mathrm{~V} & 1.0 \mathrm{~A} \\ \hline \end{array} $$

The correct choice for the parameters ( $\mathrm{I}_{\mathrm{N}}, \mathrm{R}_{\mathrm{N}}$ ) of the Norton equivalent circuit shown in Figure (b) is:

Consider the two-port network shown. For maximum power transfer to the resistive load $\left(R_L\right)$, the value of $R_L$ should be $\_\_\_\_$ $\Omega$. (Round off to two decimal places)

The resistance values of the Wheatstone bridge shown are: $P=2000 \Omega, \mathrm{q}=500 \Omega, R=3000 \Omega$. The battery voltage $E=50 \mathrm{~V}$. The battery has an internal resistance of $1 \Omega$ and the Galvanometer ( $G$ ) has a resistance of $50 \Omega$.

The value of the resistance $S$ for balanced condition is $\_\_\_\_$ $\Omega$. [Answer in integer]

A $15 \mathrm{kVA}, 1100 \mathrm{~V} / 220 \mathrm{~V}$, single-phase two-winding transformer is configured as a $1.32 \mathrm{kV} / 1.1 \mathrm{kV}$ autotransformer. What will be the rating of the autotransformer?