$A$ constant voltage of $50 \ V$ is maintained between the points $A$ and $B$ of the circuit shown in the figure. The current through the branch $CD$ of the circuit is (in $A$)

The resistance of an iron wire is $10\,\Omega$ and the temperature coefficient of resistivity is $5 \times 10^{-3}\,^{\circ}C^{-1}$. At $20\,^{\circ}C$,it carries $30\,mA$ of current. Keeping the potential difference between its ends constant,the temperature of the wire is raised to $120\,^{\circ}C$. The current in milliamperes that flows in the wire is:

In the adjacent circuit,a voltmeter of internal resistance $R$,when connected across $B$ and $C$,reads $\frac{100}{3} \text{ V}$. Neglecting the internal resistance of the battery,the value of $R$ is:

In the following circuit,if $V_A - V_B = 6\,V$,then the value of resistance $R$ (in $\Omega$) is ............... $\Omega$.

In the figure shown,battery $1$ has $emf = 6\, V$ and internal resistance $= 1\, \Omega$. Battery $2$ has $emf = 2\, V$ and internal resistance $= 3\, \Omega$. The wires have negligible resistance. What is the potential difference across the terminals of battery $2$ (in $, V$)?

The actual value of resistance $R$,shown in the figure is $30\,\Omega$. This is measured in an experiment as shown using the standard formula $R_{measured} = \frac{V}{I}$,where $V$ and $I$ are the readings of the voltmeter and ammeter,respectively. If the measured value of $R$ is $5\%$ less than the actual value,then the internal resistance of the voltmeter is ................. $\Omega$.