The image shows a simple
series circuit with the following components:
- A
10 V battery (voltage source)
- Two resistors in series:
-
R₁ = 470 Ω
-
R₂ = 1.3 kΩ (which is 1300 Ω)
We are asked to find the
total current (Iₜ) flowing through the circuit.
---
Step-by-step Solution:
#### 1.
Find Total Resistance (Rₜ)
Since the resistors are in
series, the total resistance is the sum of individual resistances:
\[
R_T = R_1 + R_2
\]
Convert R₂ to ohms:
\[
R_2 = 1.3 \, \text{kΩ} = 1300 \, \Omega
\]
Now add:
\[
R_T = 470 \, \Omega + 1300 \, \Omega = 1770 \, \Omega
\]
✔ So, total resistance:
1770 Ω
---
#### 2.
Use Ohm’s Law to Find Total Current (Iₜ)
Ohm’s Law:
\[
I_T = \frac{V_T}{R_T}
\]
Where:
- \( V_T = 10 \, \text{V} \)
- \( R_T = 1770 \, \Omega \)
\[
I_T = \frac{10}{1770} \approx 0.00565 \, \text{A}
\]
Convert to milliamps (mA):
\[
0.00565 \, \text{A} = 5.65 \, \text{mA}
\]
---
✔ Final Answer:
\[
\boxed{I_T = 5.65 \, \text{mA}}
\]
---
Additional Notes:
- Since it's a
series circuit, the same current flows through both resistors:
\( I_1 = I_2 = I_T = 5.65 \, \text{mA} \)
- You can also find the voltage drop across each resistor using:
\[
V_1 = I_T \times R_1 = 5.65 \, \text{mA} \times 470 \, \Omega \approx 2.655 \, \text{V}
\]
\[
V_2 = I_T \times R_2 = 5.65 \, \text{mA} \times 1300 \, \Omega \approx 7.345 \, \text{V}
\]
And check: \( V_1 + V_2 = 2.655 + 7.345 = 10 \, \text{V} \) → Correct!
Let me know if you'd like help finding voltages or power!
Parent Tip: Review the logic above to help your child master the concept of series circuit math.
