JEE Advance - Physics (2017 - Paper 2 Offline - No. 15)
In Process 1, the energy stored in the capacitor EC and heat dissipated across resistance ED are related by
EC = ED ln2
EC = ED
EC = 2ED
EC = $${1 \over 2}$$ED
Explanation
Energy supplied to the circuit is CV$$_0^2$$
Energy stored in capacitor, EC = $${1\over 2}$$CV$$_0^2$$
Therefore,
Energy dissipated E0 = Energy supplied $$-$$ Energy stored
= CV$$_0^2$$ $$-$$ $${1\over 2}$$CV$$_0^2$$ = $${1\over 2}$$CV$$_0^2$$
$$\Rightarrow$$ EC = ED
Energy stored in capacitor, EC = $${1\over 2}$$CV$$_0^2$$
Therefore,
Energy dissipated E0 = Energy supplied $$-$$ Energy stored
= CV$$_0^2$$ $$-$$ $${1\over 2}$$CV$$_0^2$$ = $${1\over 2}$$CV$$_0^2$$
$$\Rightarrow$$ EC = ED
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