Question

Brayden and Riku now use their skills to work a problem. Find the equivalent resistance, the current supplied by the battery and the current through each resistor when the specified resistors are connected across an 8-V battery. Part (a) uses two resistors with resistance values that can be set with the animation sliders, and you can use the animation to verify your calculation. In part (b), three resistors are specified.
(a) Two resistors are connected in series across an 8-V battery. Let R1=4 Ω and R2=1 Ω.
(b) Add a third resistor to the circuit in series. Let R1=4 Ω, R2=1 Ω, and R3=1 Ω.

Answer 1

a) $5 \Omega$, 1.6 A

b) $6 \Omega$, 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:

$R=R_1+R_2$

where

$R_1=4\Omega$

$R_2=1 \Omega$

Therefore, the equivalent resistance is

$R=4+1=5 \Omega$

Now we can use Ohm's Law to find the current flowing through the circuit:

$I=\frac{V}{R}$

where

V = 8 V is the voltage supplied by the battery

$R=5\Omega$ is the equivalent resistance of the circuit

Substituting,

$I=\frac{8}{5}=1.6 A$

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

$R=R_1+R_2+R_3$

where:

$R_1=4\Omega\\R_2=1\Omega\\R_3=1\Omega$

Substituting, we find the equivalent resistance:

$R=4+1+1=6 \Omega$

Now we can find the current through the circuit by using again Ohm's Law:

$I=\frac{V}{R}$

where

V = 8 V is the voltage supplied by the battery

$R=6\Omega$ is the equivalent resistance

Substituting,

$I=\frac{8}{6}=1.33 A$

And the three resistors are connected in series, therefore the current flowing through each resistor is the same, 1.33 A.