Question No 1 Find the voltage drop across 24 ohm resistor and current flowing through 22 ohm resistor in the given circuit as shown in figure.

8.25 V

Explanation:

We can ignore the 22Ω and 122Ω resistors at the bottom.  Since there's a short across those bottom nodes, any current will go through the short, and none through those two resistors.

The 2Ω resistor and the 44Ω resistor are in parallel.  The equivalent resistance is:

1 / (1 / (2Ω) + 1 / (44Ω)) = 1.913Ω

This resistance is in series with the 12Ω resistor.  The equivalent resistance is:

1.913Ω + 12Ω = 13.913Ω

This resistance is in parallel with the 24Ω resistor.  The equivalent resistance is:

1 / (1 / (13.913Ω) + 1 / (24Ω)) = 8.807Ω

Finally, this resistance is in series with the 4Ω resistor.  The equivalent resistance of the circuit is:

8.807Ω + 4Ω = 12.807Ω

The current through the battery is:

12 V / 12.807Ω = 0.937 A

The voltage drop across the 4Ω resistor is:

(0.937 A) (4Ω) = 3.75 V

So the voltage between the bottom nodes and the top nodes is:

12 V − 3.75 V = 8.25 V

if the thermos has a vacuum innner, then heat through conv currents and conduction are reduced = vacuum flask