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Two batteries A and B each of e.m.f. 2V are connected in series to an external resistance R = 1 Omega. If the internal resistance of battery A is 1.9 Omega and that of B is 0.9 Omega. What is the potential difference between the terminals of battery A ? |
Answer» SOLUTION :TOTAL current through the circuit `i=("voltage")/("RESISTANCE resistance")` `=(4)/((1+ 1.9+ 0.9)) =4/(3.8) A` potential difference of A, `V_A= epsi - IR`, `=2-4/(3.8)xx1.9 = 2-2 =0` |
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