Q. 12

# The 2.0 Ω resisto

Answer :

Given-

Resistance - 2.0 ΩΩ

heat capacity of the calorimeter together with water = 2000 J K–1 Looking into the circuit the effective resistance of the circuit,

2.0 Ω in parallel with 6.0 Ω and this combination in series with 1.0 Ω  Now, current i through the circuit, from ohm’s law-   Let take i’ as the current through the 6 Ω resistor.

Then, applying KCL, the algebraic sum of current entering and leaving a node is zero and from ohm’s law, we can write -      (a) Heat generated in the 2 Ω resistor, using Joule’s Heating effect from (1) and substituting values  Given that the heat capacity of the calorimeter together with water is 2000 J K−1

Which means , 2000 J of heat raise the temp by 1 K.

Then, 5832 J of heat raises the temperature of water by (b) When the 6 Ω resistor burn out, the effective resistance of the

circuit will become –  Current through the circuit, Heat generated in the 2 Ω resistor,
using Joule’s Heating effect  Given that the heat capacity of the calorimeter together with water is 2000 J K−1

Which means , 2000 J of heat raise the temp by 1 K.

So, 7200J will raise the temperature by Rate this question :

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