(a) Find the current in the 20 Ω resistor shown in figure.

(b) If a capacitor of capacitance 4 μF is joined between the points A and B, what would be the electrostatic energy stored in it in steady state?



Concepts/Formula used:


Ohm’s Law:


Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:



Kirchhoff’s junction rule:


The sum of currents entering a junction is equal to the sum of currents leaving it.


Kirchhoff’s loop rule:


The sum of potential differences around a closed loop is zero.


Energy stored by a capacitor:


If the potential difference between the two conductors of the capacitor is V and its capacitance is C, its energy is given by:


(a)



Using Kirchhoff’s law on loop FEABF,




               ........(1)


Using Kirchhoff’s law on loop ACDBA,




      ……………(2)


Solving (1) and (2), we get



Hence,



(b)



At steady state, no current passes through the capacitor; hence, the results are the same as in (a).



Now, the potential as the capacitor is in parallel with the 20Ω resistor,






Hence, the energy stored by the capacitor is 32μJ.

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