When three identical bulbs of $60 \;W, 200 \;V$ rating are connected in series to a $200 \;V$ supply,the power drawn by them will be ....... $Watt$.

In the given circuit,the current passing through the $12\, \Omega$ resistor is............ $A$.

What is the equivalent resistance between $A$ and $B$?

$A$ $100\, W$ bulb $B_1$,and two $60\, W$ bulbs $B_2$ and $B_3$,are connected to a $250\, V$ source,as shown in the figure. Now $W_1, W_2$ and $W_3$ are the output powers of the bulbs $B_1, B_2$ and $B_3$,respectively. Then

Two resistances of $400 \Omega$ and $800 \Omega$ are connected in series with a $6 \text{ V}$ battery of negligible internal resistance. $A$ voltmeter of resistance $10000 \Omega$ is used to measure the potential difference across the $400 \Omega$ resistor. The error in the measurement of potential difference in volts is approximately:

$A$ $50\,V$ battery is connected across a $10\,\Omega$ resistor. The current is $4.5\,A$. The internal resistance of the battery is ............. $\Omega$.