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Worksheet - Series Circuit Problems - Worksheet- Series Circuit ... - Free Printable

Worksheet - Series Circuit Problems - Worksheet- Series Circuit ...

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Show Answer Key & Explanations Step-by-step solution for: Worksheet - Series Circuit Problems - Worksheet- Series Circuit ...
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Let's go through each of the six series circuit problems on this worksheet and verify or solve them step by step, using Ohm’s Law and the rules for series circuits.

---

🔧 Series Circuit Rules Recap:


- Current (I) is the same through all components.
- Total Voltage (Vₜ) = Sum of voltage drops across each resistor.
- Total Resistance (Rₜ) = Sum of individual resistances:
$ R_T = R_1 + R_2 + R_3 + \dots $
- Ohm’s Law: $ V = IR $ → $ I = \frac{V}{R} $, $ V = IR $

---

## ✔ Problem 1:
Given:
- $ R_1 = 10\,\Omega $
- $ R_2 = 20\,\Omega $
- Battery voltage = 90 V

Step 1: Total Resistance
$$
R_T = R_1 + R_2 = 10 + 20 = 30\,\Omega
$$

Step 2: Total Current
$$
I_T = \frac{V_T}{R_T} = \frac{90\,V}{30\,\Omega} = 3\,A
$$

Step 3: Voltage Drops
- $ V_1 = I \cdot R_1 = 3\,A \cdot 10\,\Omega = 30\,V $
- $ V_2 = I \cdot R_2 = 3\,A \cdot 20\,\Omega = 60\,V $

✔ All values are correct:
- $ R_T = 30\,\Omega $
- $ I_T = 3\,A $
- $ I_1 = I_2 = 3\,A $
- $ V_1 = 30\,V $, $ V_2 = 60\,V $

---

## ✔ Problem 2:
Given:
- $ R_1 = 6\,\Omega $
- $ R_2 = 14\,\Omega $
- $ R_3 = 10\,\Omega $
- Battery = 60 V

Step 1: Total Resistance
$$
R_T = 6 + 14 + 10 = 30\,\Omega
$$

Step 2: Total Current
$$
I_T = \frac{60\,V}{30\,\Omega} = 2\,A
$$

Wait! The student wrote $ I_T = 3\,A $. That’s incorrect.

Let’s recalculate:
$$
I_T = \frac{60}{30} = 2\,A
$$

Then:
- $ V_1 = 2\,A \cdot 6\,\Omega = 12\,V $
- $ V_2 = 2\,A \cdot 14\,\Omega = 28\,V $
- $ V_3 = 2\,A \cdot 10\,\Omega = 20\,V $

But student wrote:
- $ V_1 = 18\,V $, $ V_2 = 42\,V $, $ V_3 = 30\,V $ — which implies current of 3 A.

✘ So student made a mistake here.

✔ Correct answers:
- $ R_T = 30\,\Omega $
- $ I_T = 2\,A $
- $ I_1 = I_2 = I_3 = 2\,A $
- $ V_1 = 12\,V $, $ V_2 = 28\,V $, $ V_3 = 20\,V $

---

## ✔ Problem 3:
Given:
- Battery = 75 V
- $ R_1 = 10\,\Omega $
- Ammeter reads 5 A → so $ I_T = 5\,A $
- Voltmeter across $ R_2 $ reads 25 V → $ V_2 = 25\,V $

Find: $ V_1 $, $ I_2 $, $ R_2 $

Since it’s a series circuit:
- $ I_T = I_1 = I_2 = 5\,A $

Voltage drop across $ R_1 $:
$$
V_1 = I \cdot R_1 = 5\,A \cdot 10\,\Omega = 50\,V
$$

Total voltage:
$$
V_T = V_1 + V_2 = 50 + 25 = 75\,V \quad \text{✓ matches battery}
$$

Now find $ R_2 $:
$$
R_2 = \frac{V_2}{I_2} = \frac{25\,V}{5\,A} = 5\,\Omega
$$

But student wrote $ R_2 = 10\,\Omega $? ✘ Incorrect.

✔ Correct:
- $ V_1 = 50\,V $
- $ I_2 = 5\,A $
- $ R_2 = 5\,\Omega $

---

## ✔ Problem 4:
Given:
- $ R_1 = 5\,\Omega $
- $ R_2 = 15\,\Omega $
- Current $ I = 5\,A $
- Find $ V_1 $, $ V_2 $, $ V_T $

Step 1: Voltage Drops
- $ V_1 = I \cdot R_1 = 5\,A \cdot 5\,\Omega = 25\,V $
- $ V_2 = I \cdot R_2 = 5\,A \cdot 15\,\Omega = 75\,V $
- $ V_T = V_1 + V_2 = 25 + 75 = 100\,V $

Student wrote:
- $ V_1 = 25\,V $
- $ V_2 = 75\,V $
- $ V_T = 100\,V $

✔ All correct!

---

## ✔ Problem 5:
Given:
- $ R_1 = 5\,\Omega $
- $ R_2 = 3\,\Omega $
- $ R_3 = 2\,\Omega $
- Battery = 20 V

Step 1: Total Resistance
$$
R_T = 5 + 3 + 2 = 10\,\Omega
$$

Step 2: Total Current
$$
I_T = \frac{20\,V}{10\,\Omega} = 2\,A
$$

Step 3: Voltage Drops
- $ V_1 = 2\,A \cdot 5\,\Omega = 10\,V $
- $ V_2 = 2\,A \cdot 3\,\Omega = 6\,V $
- $ V_3 = 2\,A \cdot 2\,\Omega = 4\,V $

Student wrote:
- $ R_T = 10\,\Omega $
- $ I_T = 2\,A $
- $ V_1 = 10\,V $, $ V_2 = 6\,V $, $ V_3 = 4\,V $

✔ All correct!

---

## ✔ Problem 6:
Given:
- Battery = 40 V
- $ R_3 = 10\,\Omega $
- Voltmeter across $ R_3 $ shows 20 V
- Voltmeter across $ R_2 $ shows 10 V
- $ I_3 = 2\,A $ (given)

We need to find $ R_1 $, $ R_2 $, $ V_1 $

Step 1: Use $ I_3 = 2\,A $


Since series: $ I_T = I_1 = I_2 = I_3 = 2\,A $

Step 2: Check $ V_3 $


$$
V_3 = I \cdot R_3 = 2\,A \cdot 10\,\Omega = 20\,V \quad \text{✓ matches voltmeter}
$$

Step 3: $ V_2 = 10\,V $ → find $ R_2 $


$$
R_2 = \frac{V_2}{I_2} = \frac{10\,V}{2\,A} = 5\,\Omega
$$

Step 4: $ V_T = 40\,V $


So $ V_1 = V_T - V_2 - V_3 = 40 - 10 - 20 = 10\,V $

Now $ R_1 = \frac{V_1}{I_1} = \frac{10\,V}{2\,A} = 5\,\Omega $

Student wrote:
- $ I_3 = 2\,A $
- $ I_1 = 2\,A $
- $ V_1 = 10\,V $
- $ R_1 = 5\,\Omega $
- $ R_2 = 5\,\Omega $

✔ All correct!

---

🟩 Summary of Corrections:



| Problem | Student Answer | Correction |
|--------|----------------|-----------|
| 1 | ✔ All correct | ✔️ |
| 2 | ✘ $ I_T = 3\,A $, voltages wrong | Should be $ I_T = 2\,A $, $ V_1=12\,V $, $ V_2=28\,V $, $ V_3=20\,V $ |
| 3 | ✘ $ R_2 = 10\,\Omega $ | Should be $ R_2 = 5\,\Omega $ |
| 4 | ✔ All correct | ✔️ |
| 5 | ✔ All correct | ✔️ |
| 6 | ✔ All correct | ✔️ |

---

✔ Final Answers (Corrected):



#### Problem 2:
- $ R_T = 30\,\Omega $
- $ I_T = 2\,A $
- $ I_1 = I_2 = I_3 = 2\,A $
- $ V_1 = 12\,V $, $ V_2 = 28\,V $, $ V_3 = 20\,V $

#### Problem 3:
- $ V_1 = 50\,V $
- $ I_2 = 5\,A $
- $ R_2 = 5\,\Omega $

All others are correct as written.

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