https://file.aiquickdraw.com/mathbot/file/1728139485605189hyb7y.jpg

The circuit in the image is a parallel combination of resistors connected to a 20V source. Here's how we can solve for each of the unknowns:

### 1. **Finding \( R_T \) (Total Resistance)**

The total resistance \( R_T \) in a parallel circuit is found using the formula:

\[
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}
\]

Where:
- \( R_1 = 5 \, \Omega \)
- \( R_2 = 2 \, \Omega \)
- \( R_3 = 7 \, \Omega \)
- \( R_4 = 8 \, \Omega \)

Substitute the values:

\[
\frac{1}{R_T} = \frac{1}{5} + \frac{1}{2} + \frac{1}{7} + \frac{1}{8}
\]

We need to compute this expression:

\[
\frac{1}{R_T} = 0.2 + 0.5 + 0.142857 + 0.125
\]

\[
\frac{1}{R_T} = 0.967857
\]

\[
R_T \approx \frac{1}{0.967857} \approx 1.032 \, \Omega
\]

So, the total resistance is approximately \( R_T = 1.032 \, \Omega \).

---

### 2. **Finding \( I_T \) (Total Current)**

The total current \( I_T \) is found using Ohm's law:

\[
I_T = \frac{V}{R_T}
\]

Where \( V = 20 \, \text{V} \) and \( R_T = 1.032 \, \Omega \).

\[
I_T = \frac{20}{1.032} \approx 19.38 \, \text{A}
\]

So, the total current is \( I_T \approx 19.38 \, \text{A} \).

---

### 3. **Finding \( V_1 \) (Voltage across each resistor)**

In a parallel circuit, the voltage across each resistor is the same and equal to the source voltage. Therefore:

\[
V_1 = V_2 = V_3 = V_4 = 20 \, \text{V}
\]

---

### 4. **Finding \( I_1 \), \( I_2 \), \( I_3 \), and \( I_4 \)**

We can use Ohm's law again to find the current through each resistor:

\[
I_1 = \frac{V}{R_1}, \quad I_2 = \frac{V}{R_2}, \quad I_3 = \frac{V}{R_3}, \quad I_4 = \frac{V}{R_4}
\]

Where \( V = 20 \, \text{V} \).

#### For \( I_1 \) (through \( R_1 = 5 \, \Omega \)):

\[
I_1 = \frac{20}{5} = 4 \, \text{A}
\]

#### For \( I_2 \) (through \( R_2 = 2 \, \Omega \)):

\[
I_2 = \frac{20}{2} = 10 \, \text{A}
\]

#### For \( I_3 \) (through \( R_3 = 7 \, \Omega \)):

\[
I_3 = \frac{20}{7} \approx 2.857 \, \text{A}
\]

#### For \( I_4 \) (through \( R_4 = 8 \, \Omega \)):

\[
I_4 = \frac{20}{8} = 2.5 \, \text{A}
\]

---

### Final Answers:

- \( R_T \approx 1.032 \, \Omega \)
- \( I_T \approx 19.38 \, \text{A} \)
- \( V_1 = V_2 = V_3 = V_4 = 20 \, \text{V} \)
- \( I_1 = 4 \, \text{A} \)
- \( I_2 = 10 \, \text{A} \)
- \( I_3 \approx 2.857 \, \text{A} \)
- \( I_4 = 2.5 \, \text{A} \)

Would you like more details on any part of the solution?

### Further Related Questions:
1. How would the total current change if one of the resistors was removed?
2. What is the power consumed by each resistor?
3. How does the total resistance change if another resistor is added in parallel?
4. What is the effect on the current if the source voltage is increased to 30V?
5. How can we find the equivalent resistance in a series circuit?

#### Tip:
In a parallel circuit, adding more resistors decreases the total resistance, while in a series circuit, it increases the total resistance.

FIND R_T=?, I_T=?, V_1=?, I_1=?, I_2=?, I_3=? for a circuit with 4 resistors in parallel: 5Ω, 2Ω, 7Ω, and 8Ω connected to a 20V source.

Learn how to find total resistance, current, and voltage in a parallel circuit with resistors of 5Ω, 2Ω, 7Ω, and 8Ω connected to a 20V source. The solution covers Ohm's Law and parallel resistance formula, suitable for students in Grades 10-12.

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