$A$ bulb rated $30\,V, 90\,W$ is to be operated on a $120\,V$ supply. What resistance (in $\Omega$) must be connected in series with it?

$A$ student was trying to construct the circuit shown in the figure below marked $(a)$,but ended up constructing the circuit marked $(b)$. Realising her mistake,she corrected the circuit,but to her surprise,the output voltage (across $R$) did not change. The value of resistance $R$ is ............ $\Omega$

$R_1$ and $R_2$ are two resistors made of different materials. The temperature coefficient of resistance for $R_1$ is $\alpha$ and for $R_2$ is $-\beta$. If the resistance of the series combination of $R_1$ and $R_2$ does not change with temperature,then the ratio of the resistances of the two wires is:

$A$ voltmeter of resistance $998\ \Omega$ is connected to a cell of $emf$ $2\ V$ and internal resistance $2\ \Omega$. The error in the measurement of $emf$ will be:

The resistances in the following circuit are $R_1 = R_2 = R_3 = 6.0 \ \Omega$. The emf of the battery is $12 \ V$. When switch $S$ is closed, the potential across resistance $R_1$ changes by an amount:

$A$ straight wire of resistance $R$ is bent into the shape of a square. $A$ cell of emf $12 \text{ V}$ is connected between two adjacent corners of the square. The potential difference across any diagonal of the square is (in $\text{ V}$)