If two bulbs of wattage $25\, W$ and $30\, W$,each rated at $220\, V$,are connected in series with a $440\, V$ supply,which bulb will fuse?

The ammeter $A$ reads $2\, A$ and the voltmeter $V$ reads $20\, V$. The value of resistance $R$ is (Assuming finite resistances of ammeter and voltmeter).

The series combination of two batteries,both of the same emf $10 \; V,$ but different internal resistances of $20 \; \Omega$ and $5 \; \Omega,$ is connected to the parallel combination of two resistors $30 \; \Omega$ and $R \; \Omega.$ If the voltage difference across the battery of internal resistance $20 \; \Omega$ is zero,the value of $R$ (in $\Omega$) is:

When an ammeter of negligible internal resistance is inserted in series with the circuit,it reads $1 \, A$. When a voltmeter of very large resistance is connected across $X$,it reads $1 \, V$. When points $A$ and $B$ (across $Y$) are shorted by a conducting wire,the voltmeter measures $10 \, V$ across the battery. The internal resistance of the battery is equal to .............. $\Omega$.

Two identical bulbs,each rated $500 \, W$ and $220 \, V$,are connected in series to a $110 \, V$ supply. What is the power consumed by each bulb?

$A$ current of $1.5\, A$ flows through a copper voltameter. The thickness of copper deposited on the electrode surface of area $50\, cm^2$ in $20\, min$ will be (Density of copper $= 9000\, kg/m^3$ and $E.C.E.$ of copper $= 0.00033\, g/C$).