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Two identical calls send the same current in `2 Omega` resistance, whether connected in series of in parallel. The internal resistance of the cell should beA. `1 Omega`B. `2 Omega`C. `(1)/(2) Omega`D. `2.5 Omega` |
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Answer» Correct Answer - B (b) In series, `i_(1) = (2E)/(2 + 2r)` In parallel, `I_(2) = (E)/(2 + (r )/(2)) = (2E)/(4 + r)` Since `i_(1) = i_(2) implies (2E)/(4 + r) = (2E)/(2 + 2r) implies r = 2 Omega` |
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