Give step-by-step solution with explanation and final answer:Electric circuits form the foundation of electrical engineering, as all
machines and systems ultimately rely on current and volfage behaiv

Question

Give step-by-step solution with explanation and final answer:Electric circuits form the foundation of electrical engineering, as all
machines and systems ultimately rely on current and volfage behaivior
in resistive, inductive, or capacitive eiements. lniderstanding simple
resistive circuits is the first step toward mastering more advanced elec-
trical networks. The ability to apply Ohm's Law and basic power form-
ulas is essential for engineers to analyze ad design practical systems
such as lighting circuits, battery chargers, and household appliances.
A DC circuit consists of a 12 V battery connected across a resistor of 6 Q.
This is one of the simplest examples of an electrical network, yet it dem-
onstrates the basic consepts of current flow, voltage drop, and power
dissipation. Engineers often begin with such elementary circuits to
understand how electrical energy is converted into heat or
Given:
+ Battery Voltage, V=12V
» Resistor, R=6Q
Determine:
1. The current flowing through the resistor.
2. Formula for current: I = z
3. Substitution of values to calculate current:

P=VxIor P=I*R
4. Substitution of calculated current to find power.
6. Final numerical

VIPSolutions · Accepted Answer

Let's solve the problem step by step:

---

## **Given**
- Battery Voltage, $ V = 12 $ V  
- Resistor, $ R = 6 $ Ω

---

## **1. Find the Current through the Resistor**

**Formula:**  
$$
I = \frac{V}{R}
$$

**Substitute the values:**  
$$
I = \frac{12\, \text{V}}{6\, \Omega}
$$

$$
I = 2\, \text{A}
$$

**So, the current flowing through the resistor is:**  
$$
\boxed{2\, \text{A}}
$$

---

## **2. Find the Power Dissipated in the Resistor**

**Formula (using $ P = V \times I $ or $ P = I^2 R $)**  
Let's use $ P = V \times I $:

$$
P = 12\, \text{V} \times 2\, \text{A}
$$
$$
P = 24\, \text{W}
$$

Or, using $ P = I^2 R $:

$$
P = (2\, \text{A})^2 \times 6\, \Omega
$$
$$
P = 4 \times 6
$$
$$
P = 24\, \text{W}
$$

**So, the power dissipated is:**  
$$
\boxed{24\, \text{W}}
$$

---

## **Final Answers**

- **Current through the resistor:** $ \boxed{2\, \text{A}} $
- **Power dissipated in the resistor:** $ \boxed{24\, \text{W}} $

---

**Summary Table:**

| Quantity | Formula | Value |
|----------|---------|-------|
| Current ($I$) | $I = \frac{V}{R}$ | $2\, \text{A}$ |
| Power ($P$) | $P = V \times I$ or $P = I^2 R$ | $24\, \text{W}$ |

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