Let's analyze the circuit shown in your image and solve for the unknown values, specifically
V₃,
I₃, and
R₃.
---
🔧 Given:
- Total voltage: $ V_T = 12 \text{V} $
- Total current: $ I_T = 2 \text{A} $
- Total resistance: $ R_T = 6\ \Omega $
We have three resistors in
series (since the same current flows through all of them):
- $ R_1 = 3\ \Omega $, $ V_1 = 6\ \text{V} $, $ I_1 = 2\ \text{A} $
- $ R_2 = 2\ \Omega $, $ V_2 = 4\ \text{V} $, $ I_2 = 2\ \text{A} $
- $ R_3 = ? $, $ V_3 = ? $, $ I_3 = ? $
---
✔ Step 1: Confirm Series Configuration
Since:
- The current is the same throughout: $ I_1 = I_2 = I_3 = I_T = 2\ \text{A} $
- The total voltage is the sum of individual voltages
- Resistors are connected end-to-end with no branching
→ This is a
series circuit
---
✔ Step 2: Use Kirchhoff’s Voltage Law (KVL)
In a series circuit:
$$
V_T = V_1 + V_2 + V_3
$$
Substitute known values:
$$
12\ \text{V} = 6\ \text{V} + 4\ \text{V} + V_3
$$
$$
V_3 = 12 - 6 - 4 = 2\ \text{V}
$$
✔ So, $ \boxed{V_3 = 2\ \text{V}} $
---
✔ Step 3: Current through $ R_3 $
In series, current is constant:
$$
I_3 = I_T = 2\ \text{A}
$$
✔ So, $ \boxed{I_3 = 2\ \text{A}} $
---
✔ Step 4: Find $ R_3 $
Use Ohm’s Law:
$$
R_3 = \frac{V_3}{I_3} = \frac{2\ \text{V}}{2\ \text{A}} = 1\ \Omega
$$
✔ So, $ \boxed{R_3 = 1\ \Omega} $
---
✔ Verify Total Resistance
Check if $ R_T = R_1 + R_2 + R_3 $
$$
R_T = 3\ \Omega + 2\ \Omega + 1\ \Omega = 6\ \Omega
$$
Matches the given $ R_T $. ✔️
Also verify using Ohm’s Law:
$$
R_T = \frac{V_T}{I_T} = \frac{12\ \text{V}}{2\ \text{A}} = 6\ \Omega
$$
✔️ Confirmed.
---
🟩 Final Answers:
| Quantity | Value |
|--------|-------|
| $ V_3 $ | $ 2\ \text{V} $ |
| $ I_3 $ | $ 2\ \text{A} $ |
| $ R_3 $ | $ 1\ \Omega $ |
---
✔ Summary:
The missing values are:
- $ \boxed{V_3 = 2\ \text{V}} $
- $ \boxed{I_3 = 2\ \text{A}} $
- $ \boxed{R_3 = 1\ \Omega} $
All values are consistent with series circuit laws and Ohm’s Law.
Parent Tip: Review the logic above to help your child master the concept of series circuit math.
