Question

You are given two circuits with two batteries of emf and internalresistance $R_1$ each. Circuit A has the batteries connected inserieswith a resistor of resistance $R_2$ , and circuit B has thebatteries connected in parallel toan equivalent resistor.Note that the symbol should be entered in your answersasEMF.
$15092_a.jpg$ Calculate the current $I_B$ through the resistor ofresistance $R_2$ for circuit B.Express your answer in terms of, $R_1$ , and $R_2$ .what I get: $I_A$ = $left(2{cal{E}}right)/left(R_{2}+2R_{1}right)$ but what the heckis $I_B$ ?

Answer 1

Modified form ofrcuit AOhm'slawis V=IRI=V/RResultentresistance R=2R₁R₂/(2R₁+R₂)I=2E(2R₁+R₂)/2R₁R₂Modified form of circuit B isSince there is no voltage source no current will flow through thiscircuit.