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In the given circuit, with steady current, the potential drop across the capacitor must be A. VB. `V//2`C. `V//3`D. `2V//3` |
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Answer» Correct Answer - C V = E -ir. When I = 0, the potential reading is 2V. Hence emf is 2 V. When V = 0, I = 5A. Thisk gives `r = 0.4 Omega.` |
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