Answer to Question #276684 in Electric Circuits for Librio

Question

An electric lamp whose resistance, when in use, is 2 Ω is connected to the terminals of a dry cell whose e.m.f. is 1.5 V. If the current through the lamp is 0.5 A, calculate the internal resistance of the cell and the potential difference between the terminals of the lamp. If two such cells are connected in parallel, find the resistance which must be connected in series with the arrangement to keep the current the same as before.

Answer 1

$r=(E-IR)/I=(1.5-0.5\cdot2)/0.5=1\ (\Omega)$

$U=E-Ir=1.5-0.5\cdot1=1\ (V)$

$0.5=1.5/(2+0.5+x)\to x=0.5 (\Omega)$