Question

part a
determine the magnitude of the current through $r_1$ in the figure.(figure 1)
express your answer to two significant figures and include the appropriate units.
$v_1 = 9.0v$ $r_1 = 22omega$ $r_2 = 18omega$ $v_3 = 6.0v$
part b
determine the direction of the current through $r_1$ in the figure.

Explanation:

Step1: Apply Kirchhoff's laws and Ohm's law

We use the loop - rule. Let's assume the current through $R_1$ is $I_1$, through $R_2$ is $I_2$. For the left - hand loop with $V_1$ and $R_1$, and the right - hand loop with $V_3$ and $R_2$, and considering the junction rule. But we can also use the superposition principle or directly analyze the circuit. The voltage across $R_1$ is determined by the net emf in the circuit. The net emf driving the current through $R_1$ is $V = V_1=9.0\ V$ (since there is no other emf in the path of the current through $R_1$ in a simple sense). According to Ohm's law $I=\frac{V}{R}$, where $V$ is the voltage across the resistor and $R$ is the resistance.

Step2: Calculate the current through $R_1$

Using Ohm's law $I_1=\frac{V_1}{R_1}$. Substitute $V_1 = 9.0\ V$ and $R_1=22\ \Omega$ into the formula: $I_1=\frac{9.0}{22}\ A\approx0.41\ A$.

The positive terminal of the battery $V_1$ is the higher - potential end. Current flows from higher potential to lower potential in a circuit. So the current through $R_1$ in the direction indicated in the circuit (from left to right) is the correct direction.

Answer:

$I_1 = 0.41\ A$

For Part B:
The current through $R_1$ is from the positive terminal of $V_1$ to the negative terminal, so the direction is to the right.