Question

A battery consisting of three identical cells, each with internal resistance of 1.0 Ω is connected in a series with

5.0 Ω and 3.0 Ω resistors. The voltage across the 5 Ω resistor is 5V. Determine the (a) current flowing through the circuit, (b) voltage across the 3 Ω resistor, (C) terminal voltage of the cell, and (d) electromotive force of the cell.

Answer 1

Solution.

$r_1=1.0 Om;$

$R_1=5.0 Om;$

$V_1=5.0V;$

$R_2=3.0Om;$

$a) I=\dfrac{V_1}{R_1};$

$I=\dfrac{5.0}{5.0}=1.0A;$

$b)V_2=IR_2;$

$V_2=1.0\sdot3.0=3.0V;$

$c)R=R_1+R_2;$

$R=5.0+3.0=8.0Om;$

$V=IR;$

$V=1.0\sdot8.0=8.0V;$

$d) EMF=I(R+r); r=3r_1;$

$EMF=1.0(8.0+3.0)=11V;$

$EMF_1=\dfrac{EMF}{3};$

$EMF_1=\dfrac{11}{3}=3.7V;$

Answer: $a) I=1.0A;$

$b)V_2=3.0V;$

$c)V=8.0V;$

$d)EMF_1=3.7V.$