Answer:
0.2 A
Explanation:
First of all, let's find the total resistance of the two resistors in parallel:
[tex]\frac{1}{R_{12}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{15 \Omega}+\frac{1}{30 \Omega}=\frac{3}{30 \Omega}\\R_{12}=\frac{30}{3}\Omega = 10 \Omega[/tex]
This combination is then connected in series with the other resistor, so the total resistance of the circuit is
[tex]R_T = R_{12}+R_3 = 10 \Omega + 20 \Omega = 30 \Omega[/tex]
Then we can calculate the total current in the circuit
[tex]I=\frac{V}{R_T}=\frac{9.0 V}{30 \Omega}=0.3 A[/tex]
This current flows through the resistor of 20-Ω, then it splits in the two branches of the 15-Ω and 30-Ω resistors.
The voltage drop across the resistor of 20-Ω is
[tex]V_3 = IR_3 = (0.3 A)(20 \Omega)=6 V[/tex]
So, the potential difference across the two resistors in parallel is
[tex]V_{12}=V-V_3 = 9 V - 6 V= 3V[/tex]
So, the current through the 15-Ω resistor is
[tex]I_1 = \frac{V_{12}}{R_1}=\frac{3.0 V}{15 \Omega}=0.2 A[/tex]
