The series combination of two batteries,both | vedclass

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(A) Let the current in the circuit be $i.$ The batteries are in series,so the total emf is $E_{eq} = 10 + 10 = 20 \; V$ and the total internal resista

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) Let the current in the circuit be $i.$ The batteries are in series,so the total emf is $E_{eq} = 10 + 10 = 20 \; V$ and the total internal resistance is $r_{eq} = 20 + 5 = 25 \; \Omega.$The voltage across the battery with internal resistance $20 \; \Omega$ is given by $V_1 = E - i r_1 = 10 - i(20).$Given $V_1 = 0,$ we have $10 - 20i = 0,$ which gives $i = 0.5 \; A.$The total external resistance $R_{ext}$ is the parallel combination of $30 \; \Omega$ and $R \; \Omega,$ so $R_{ext} = \frac{30R}{30+R}.$Using Ohm's law for the whole circuit,$E_{eq} = i(R_{ext} + r_{eq}).$$20 = 0.5 \left( \frac{30R}{30+R} + 25 \right).$$40 = \frac{30R}{30+R} + 25.$$15 = \frac{30R}{30+R}.$$15(30+R) = 30R.$$450 + 15R = 30R.$$15R = 450 \implies R = 30 \; \Omega.$

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