Question

1. Consider the circuit shown in Figure 1. Complete the table below for that circuit. Equivalent R of circuit _ This circuit is a _ divider (i.e.. what quantity is divide up between the resistors?).

Answer 1

From information given we have the current in series circuit so

$I_1 = I_2 =I_{battery} = 0.2 A$

for R1

$\Delta V_{1} = I_1 R_1 = (0.2)(10 ) = 2V$

the voltage for battery:
$\Delta V= \Delta V_1 + \Delta V_2 = 2+5 = 7V$

to know R2 we use the Ohm's law:

$R_2 = \Delta V_2 /I_2 = (5)(0.2) = 25 \Omega$

To know the power delivered we use:

$P_{battery} =+ VI = (7)(0.2) = +1.4 W$

$P_{R_1} =-I_1^2 R_1= (0.2)^2(10) = -0.4 W$

$P_{R_2} =-I_2^2 R_2= (0.2)^2(25) = -1 W$

The equivaelnte resistance is R = R1 + R2 = 35

the circuit is voltage divide