The image shows a
simple series circuit consisting of:
- A
20 V battery (voltage source)
- Three resistors in series:
- 20 Ω
- 30 Ω
- 40 Ω
We are likely being asked to find either:
- The total current in the circuit
- The voltage drop across each resistor
- Or the equivalent resistance
Let’s solve this step by step.
---
Step 1: Find the total resistance
Since all resistors are in
series, we add their resistances:
\[
R_{\text{total}} = 20\,\Omega + 30\,\Omega + 40\,\Omega = 90\,\Omega
\]
---
Step 2: Use Ohm's Law to find the current
Ohm’s Law:
\[
I = \frac{V}{R}
\]
Where:
- \( V = 20\, \text{V} \)
- \( R = 90\, \Omega \)
\[
I = \frac{20\, \text{V}}{90\, \Omega} = \frac{2}{9}\, \text{A} \approx 0.222\, \text{A}
\]
So, the
current through the circuit is approximately 0.222 A (or exactly \( \frac{2}{9} \) A).
---
Step 3: Voltage drops across each resistor
Using Ohm’s Law again: \( V = I \times R \)
- Across 20 Ω:
\[
V_1 = I \times 20 = \frac{2}{9} \times 20 = \frac{40}{9} \approx 4.44\, \text{V}
\]
- Across 30 Ω:
\[
V_2 = \frac{2}{9} \times 30 = \frac{60}{9} \approx 6.67\, \text{V}
\]
- Across 40 Ω:
\[
V_3 = \frac{2}{9} \times 40 = \frac{80}{9} \approx 8.89\, \text{V}
\]
Check: Total voltage drop =
\[
4.44 + 6.67 + 8.89 = 20\, \text{V} \quad \text{(matches the battery voltage)}
\]
---
✔ Final Answers:
-
Total resistance: \( 90\, \Omega \)
-
Current: \( \frac{2}{9} \, \text{A} \approx 0.222\, \text{A} \)
-
Voltage drops:
- 20 Ω: ~4.44 V
- 30 Ω: ~6.67 V
- 40 Ω: ~8.89 V
If the task was to find the
current, that’s the key result.
Let me know if you were looking for something specific like power or another quantity!
Parent Tip: Review the logic above to help your child master the concept of series circuit math.
