In the circuit shown in the given figure,the resistances $R_1$ and $R_2$ are respectively:

The value of the resistance $R$ in the figure is .......... $\Omega$.

Five current-carrying conductors meet at point $P$. What is the magnitude and direction of the current in conductor $PQ$?

The potential difference $V$ across the filament of the bulb shown in the given Wheatstone bridge varies as $V = i(2i + 1)$,where $i$ is the current in ampere through the filament of the bulb. The emf of the battery $(V_b)$ so that the bridge becomes balanced is (in $V$)

The four arms of a Wheatstone bridge (Figure) have the following resistances: $AB = 100 \; \Omega$,$BC = 10 \; \Omega$,$CD = 5 \; \Omega$,and $DA = 60 \; \Omega$. $A$ galvanometer of $15 \; \Omega$ resistance is connected across $BD$. Calculate the current through the galvanometer when a potential difference of $10 \; V$ is maintained across $AC$.

What are the advantages of the null-point method in a Wheatstone bridge? What additional measurements would be required to calculate $R_{\text{unknown}}$ by any other method?