Answer:
11.3 A
Explanation:
First of all, we need to find the equivalent resistance of the two appliances in parallel, which is given by:
[tex]\frac{1}{R_{12}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{15 \Omega}+\frac{1}{20 \Omega}=0.116 \Omega^{-1}\\R_{12} = 8.6 \Omega[/tex]
Then we know that these two appliances are connected in series to another resistor of
[tex]R_3 = 2 \Omega[/tex]
So the total resistance of the circuit is
[tex]R=R_{12}+R_3 = 8.6\Omega + 2\Omega = 10.6\Omega[/tex]
And since we know the voltage of the source,
V = 120 V
We can now calculate the current in the circuit by using Ohm's law:
[tex]I=\frac{V}{R}=\frac{120 V}{10.6 \Omega}=11.3 A[/tex]
