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.
$v_1 = 9.0v$ $r_1 = 22omega$ $r_2 = 18omega$ $v_3 = 6.0v$
$i_1=$ value units
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part b
determine the direction of the current through $r_1$ in the figure.

Explanation:

Step1: Apply Kirchhoff's laws

We can use Kirchhoff's voltage - law (KVL) to analyze the loop. Let's assume the current through $R_1$ is $I_1$ and the current through $R_2$ is $I_2$. For the outer loop, starting from the negative terminal of $V_1$ and moving clock - wise, we have $V_1 - I_1R_1-I_2R_2 = 0$. For the inner loop with $V_3$, starting from the negative terminal of $V_3$ and moving clock - wise, we have $V_3 - I_2R_2=0$.

Step2: Solve for $I_2$ from the inner - loop equation

From $V_3 - I_2R_2 = 0$, we can express $I_2$ as $I_2=\frac{V_3}{R_2}$. Substituting $V_3 = 6.0\ V$ and $R_2 = 18\ \Omega$, we get $I_2=\frac{6.0}{18}=\frac{1}{3}\ A$.

Step3: Substitute $I_2$ into the outer - loop equation and solve for $I_1$

Substitute $I_2$ into $V_1 - I_1R_1-I_2R_2 = 0$. We know $V_1 = 9.0\ V$, $R_1 = 22\ \Omega$, and $R_2 = 18\ \Omega$, $I_2=\frac{1}{3}\ A$. So $9.0-22I_1 - 18\times\frac{1}{3}=0$. Simplify the equation: $9.0-22I_1 - 6.0 = 0$, which gives $3.0-22I_1 = 0$. Then $22I_1=3.0$, and $I_1=\frac{3.0}{22}\approx0.14\ A = 140\ \mu A$.

We assume the direction of the current based on the polarity of the voltage sources. Since $V_1$ has a higher voltage and its positive terminal is on the right - hand side of the loop, the current through $R_1$ is to the right.

Answer:

$I_1 = 140\ \mu A$

For part B: