a) [tex]5 \Omega[/tex], 1.6 A
b) [tex]6 \Omega[/tex], 1.33 A
Explanation:
a)
In this situation, we have two resistors connected in series.
The equivalent resistance of resistors in series is equal to the sum of the individual resistances, so in this circuit:
[tex]R=R_1+R_2[/tex]
where
[tex]R_1=4\Omega[/tex]
[tex]R_2=1 \Omega[/tex]
Therefore, the equivalent resistance is
[tex]R=4+1=5 \Omega[/tex]
Now we can use Ohm's Law to find the current flowing through the circuit:
[tex]I=\frac{V}{R}[/tex]
where
V = 8 V is the voltage supplied by the battery
[tex]R=5\Omega[/tex] is the equivalent resistance of the circuit
Substituting,
[tex]I=\frac{8}{5}=1.6 A[/tex]
The two resistors are connected in series, therefore the current flowing through each resistor is the same, 1.6 A.
b)
In this part, a third resistor is added in series to the circuit; so the new equivalent resistance of the circuit is
[tex]R=R_1+R_2+R_3[/tex]
where:
[tex]R_1=4\Omega\\R_2=1\Omega\\R_3=1\Omega[/tex]
Substituting, we find the equivalent resistance:
[tex]R=4+1+1=6 \Omega[/tex]
Now we can find the current through the circuit by using again Ohm's Law:
[tex]I=\frac{V}{R}[/tex]
where
V = 8 V is the voltage supplied by the battery
[tex]R=6\Omega[/tex] is the equivalent resistance
Substituting,
[tex]I=\frac{8}{6}=1.33 A[/tex]
And the three resistors are connected in series, therefore the current flowing through each resistor is the same, 1.33 A.
