A) 0.05 J and 0 m J
B) 0.05 mJ and 0.05 m J
C) 0.4 J and 0.2 J
D) 1.0 m J and 0.5 m J
Correct Answer: C
Solution :
Energy drawn from battery = potential difference (V) \[\times \] charge (q) Also, charge (q) = capacitance \[\times \] potential difference (V) \[\therefore \] \[E=V\times C\,V=C{{V}^{2}}\] Given, \[C=10\,\mu F=10\times {{10}^{-6}}F\], V = 200 volt \[\therefore \] \[E=(10\times {{10}^{-6}})\,{{(200)}^{2}}=0.4\,J\] Work done in charging the conductor is stored as potential energy in the electric field, given by \[U=\frac{1}{2}C{{V}^{2}}\] Given, \[C=10\,\mu F=10\times {{10}^{-6}}F\], V = 200 volt \[U=\frac{1}{2}\times 10\times {{10}^{-6}}\times {{(200)}^{2}}\] = 0.2 J
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