the 10.00 v battery in the figure (figure 1) is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. the rest of the circuit is as shown in the figure.
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figure
part a
find the current through 3.00 ω resistor.
express your answer with the appropriate units.
part b
find the current through 4.00 ω resistor.
express your answer with the appropriate units.

Question

Accepted Answer

# Explanation:
## Step1: Apply Kirchhoff's Voltage Law (KVL)
For the left - hand loop (with the $5.00\ V$ battery and $4.00\ \Omega$ resistor), assume the current through the $4.00\ \Omega$ resistor is $I_1$ (downward) and for the right - hand loop (with the $10.00\ V$ battery and $3.00\ \Omega$ resistor), assume the current through the $3.00\ \Omega$ resistor is $I_2$ (downward).
For the left loop: $5 - 4I_1-10I_{total}=0$ (where $I_{total}$ is the current through the $10\ \Omega$ resistor). For the right loop: $10 - 3I_2-10I_{total}=0$. Also, by Kirchhoff's Current Law (KCL), $I_{total}=I_1 + I_2$.

## Step2: Solve the system of equations
From the left - loop equation: $I_1=\frac{5 - 10I_{total}}{4}$. From the right - loop equation: $I_2=\frac{10 - 10I_{total}}{3}$. Substitute into $I_{total}=I_1 + I_2$:
$$
\begin{align*}
I_{total}&=\frac{5 - 10I_{total}}{4}+\frac{10 - 10I_{total}}{3}\
12I_{total}&=3(5 - 10I_{total})+4(10 - 10I_{total})\
12I_{total}&=15-30I_{total}+40 - 40I_{total}\
12I_{total}+30I_{total}+40I_{total}&=55\
82I_{total}&=55\
I_{total}&=\frac{55}{82}\approx0.671\ A
\end{align*}
$$

## Step3: Calculate the current through the $3.00\ \Omega$ resistor (Part A)
Using $10 - 3I_2-10I_{total}=0$, substitute $I_{total}=\frac{55}{82}$:
$$
\begin{align*}
10-3I_2 - 10\times\frac{55}{82}&=0\
3I_2&=10-\frac{550}{82}\
3I_2&=\frac{820 - 550}{82}\
3I_2&=\frac{270}{82}\
I_2&=\frac{90}{82}\approx1.098\ A = 1.10\times10^{3}\ mA
\end{align*}
$$

## Step4: Calculate the current through the $4.00\ \Omega$ resistor (Part B)
Using $5 - 4I_1-10I_{total}=0$, substitute $I_{total}=\frac{55}{82}$:
$$
\begin{align*}
5-4I_1-10\times\frac{55}{82}&=0\
4I_1&=5-\frac{550}{82}\
4I_1&=\frac{410 - 550}{82}\
4I_1&=-\frac{140}{82}\
I_1&=-\frac{35}{82}\approx - 0.427\ A
\end{align*}
$$
The magnitude of the current through the $4.00\ \Omega$ resistor is $0.427\ A = 427\ mA$

# Answer:
Part A: $1.10\times 10^{3}\ mA$
Part B: $427\ mA$

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QUESTION IMAGE

Question

Explanation:

Step1: Apply Kirchhoff's Voltage Law (KVL)

Step2: Solve the system of equations

Step3: Calculate the current through the \(3.00\ \Omega\) resistor (Part A)

Step4: Calculate the current through the \(4.00\ \Omega\) resistor (Part B)

Answer: