Find the equivalent resistance between points $A$ and $B$ in the network of resistors,each of resistance $r \Omega$.

Four resistances $40 \ \Omega, 60 \ \Omega, 90 \ \Omega$,and $110 \ \Omega$ form the arms of a quadrilateral $ABCD$. $A$ battery of emf $40 \ V$ with negligible internal resistance is connected across $AC$. Find the potential difference across $BD$. (in $V$)

What is the current $I$ (in $A$) flowing through the given circuit?

Six resistors of $3 \; \Omega$ each are connected along the sides of a hexagon and three resistors of $6 \; \Omega$ each are connected along $AC$, $AD$, and $AE$ as shown in the figure. The equivalent resistance between $A$ and $B$ is equal to

In the circuit given, $E = 6.0 \, V$, $R_1 = 100 \, \Omega$, $R_2 = R_3 = 50 \, \Omega$, and $R_4 = 75 \, \Omega$. The equivalent resistance of the circuit, in $\Omega$, is:

Two identical bulbs are connected in parallel across an ideal source of emf $E$. The ammeter $A$ and voltmeter $V$ are ideal. If bulb $B_2$ gets fused,then