R49 · Ohm's law: resistance and power dissipation

To solve this problem, we can use Ohm's law and the formula for electrical power.

(a) **Calculating the Resistance:**

Ohm's law states that \( V = IR \), where:
- \( V \) is the voltage,
- \( I \) is the current, and
- \( R \) is the resistance.

Given:
- \( V = 12 \) volts,
- \( I = 0.5 \) amperes.

We can rearrange Ohm's law to solve for \( R \):
\[ R = \frac{V}{I} \]

Substituting the given values:
\[ R = \frac{12 \, \text{V}}{0.5 \, \text{A}} \]
\[ R = 24 \, \Omega \]

So, the resistance of the resistor is \( 24 \, \Omega \).

(b) **Calculating the Electrical Power:**

The formula for electrical power dissipated in a resistor is:
\[ P = VI \]

Given:
- \( V = 12 \) volts,
- \( I = 0.5 \) amperes.

Substituting the given values:
\[ P = 12 \, \text{V} \times 0.5 \, \text{A} \]
\[ P = 6 \, \text{W} \]

So, the electrical power dissipated in the resistor is \( 6 \, \text{W} \).

### Summary:
(a) The resistance of the resistor is \( 24 \, \Omega \).
(b) The electrical power dissipated in the resistor is \( 6 \, \text{W} \).

To solve this problem, we will use Ohm's law and the formula for electrical power.

**Given values:**
*   Voltage ($V$) = $12\text{ V}$
*   Current ($I$) = $0.5\text{ A}$

---

### (a) Calculation of Resistance ($R$)
According to Ohm's law, the relationship between voltage, current, and resistance is defined as $V = I \times R$. To find the resistance, we rearrange the formula:

$$R = \frac{V}{I}$$

**Calculation:**
$$R = \frac{12\text{ V}}{0.5\text{ A}}$$
$$R = 24\ \Omega$$

**The resistance of the resistor is $24\ \Omega$.**

---

### (b) Calculation of Electrical Power ($P$)
The electrical power dissipated in a circuit can be calculated using the formula $P = V \times I$.

**Calculation:**
$$P = 12\text{ V} \times 0.5\text{ A}$$
$$P = 6\text{ W}$$

*(Alternatively, using $P = I^2 \times R$: $P = (0.5\text{ A})^2 \times 24\ \Omega = 0.25 \times 24 = 6\text{ W}$)*

**The electrical power dissipated is $6\text{ W}$.**

(a) To find the resistance $ R $ of the resistor using Ohm's law $ V = IR $, rearrange the formula to solve for $ R $:

$$
R = \frac{V}{I} = \frac{12\, \text{V}}{0.5\, \text{A}} = 24\, \Omega
$$

**Resistance of the resistor:** $ \boxed{24\, \Omega} $

---

(b) To calculate the electrical power $ P $ dissipated in the resistor, use the formula $ P = VI $:

$$
P = V \times I = 12\, \text{V} \times 0.5\, \text{A} = 6\, \text{W}
$$

**Electrical power dissipated:** $ \boxed{6\, \text{W}} $

Given a 12 V battery connected across a resistor with a steady current of 0.5 A, we can use Ohm's law and power formulas to find the required values.

(a) Using Ohm's law \( V = IR \), we rearrange to solve for resistance:
\[ R = \frac{V}{I} = \frac{12 \, \text{V}}{0.5 \, \text{A}} = 24 \, \Omega \]

(b) To find the electrical power dissipated, we use the formula \( P = VI \):
\[ P = V \times I = 12 \, \text{V} \times 0.5 \, \text{A} = 6 \, \text{W} \]

Alternatively, using other power formulas confirms the result:
- Using \( P = I^2R \):
  \[ (0.5 \, \text{A})^2 \times 24 \, \Omega = 0.25 \, \text{A}^2 \times 24 \, \Omega = 6 \, \text{W} \]
- Using \( P = \frac{V^2}{R} \):
  \[ \frac{(12 \, \text{V})^2}{24 \, \Omega} = \frac{144 \, \text{V}^2}{24 \, \Omega} = 6 \, \text{W} \]

Both methods confirm the power dissipated is 6 watts.

### Final Answer

(a) The resistance of the resistor is \boxed{24} ohms.

(b) The electrical power dissipated in it is \boxed{6} watts.

## Step 1: Identify the given information
We are given a 12 V battery connected across a resistor, resulting in a steady current of 0.5 A through the circuit.

## Step 2: Apply Ohm's law to calculate the resistance
According to Ohm's law, R = V/I, where R is the resistance, V is the voltage, and I is the current. We can plug in the given values: R = 12 V / 0.5 A.

## Step 3: Perform the calculation for resistance
R = 12 V / 0.5 A = 24 ohms.

## Step 4: Calculate the electrical power dissipated in the resistor
The electrical power (P) dissipated in the resistor can be calculated using the formula P = V * I, where V is the voltage and I is the current.

## Step 5: Perform the calculation for power
P = 12 V * 0.5 A = 6 watts.

The final answer is: $\boxed{24}$

Here are the calculations using Ohm's Law and the power formulas.

### Given Values:
* Voltage ($V$) = $12 \, \text{V}$
* Current ($I$) = $0.5 \, \text{A}$

---

### (a) Calculate the resistance of the resistor ($R$).

Ohm's Law states: $V = I \times R$
To find the resistance, we rearrange the formula: $R = \frac{V}{I}$

**Calculation:**
$$R = \frac{12 \, \text{V}}{0.5 \, \text{A}}$$
$$R = 24 \, \Omega$$

**Answer (a): The resistance of the resistor is $\mathbf{24 \, \Omega}$.**

---

### (b) Calculate the electrical power dissipated in it ($P$).

The formula for electrical power when voltage and current are known is: $P = V \times I$

*(Alternatively, using the result from part (a): $P = I^2 R = (0.5 \, \text{A})^2 \times 24 \, \Omega = 0.25 \times 24 = 6 \, \text{W}$)*

**Calculation:**
$$P = 12 \, \text{V} \times 0.5 \, \text{A}$$
$$P = 6 \, \text{W}$$

**Answer (b): The electrical power dissipated is $\mathbf{6 \, \text{W}}$.**

### Given Values:
*   Voltage ($V$) $= 12 \text{ V}$
*   Current ($I$) $= 0.5 \text{ A}$

***

### (a) Calculate the resistance of the resistor

We use Ohm's Law, which states that $V = I \times R$. To find the resistance ($R$), we rearrange the formula:
$$R = \frac{V}{I}$$

**Calculation:**
$$R = \frac{12 \text{ V}}{0.5 \text{ A}}$$
$$R = 24 \text{ } \Omega$$

**(a) The resistance of the resistor is $\mathbf{24 \text{ } \Omega}$.**

***

### (b) Calculate the electrical power dissipated in it

We use the formula for electrical power, $P = V \times I$.

**Calculation:**
$$P = 12 \text{ V} \times 0.5 \text{ A}$$
$$P = 6 \text{ W}$$

*(Alternatively, using the calculated resistance: $P = I^2 R = (0.5 \text{ A})^2 \times 24 \text{ } \Omega = 0.25 \times 24 = 6 \text{ W}$)*

**(b) The electrical power dissipated is $\mathbf{6 \text{ W}}$.**

Here is the step-by-step calculation using Ohm's Law and the power formulas.

### Given Information:
* Voltage ($\text{V}$) = $12\ \text{V}$
* Current ($I$) = $0.5\ \text{A}$

---

### (a) Resistance of the Resistor ($R$)

Ohm's Law states that $V = I \times R$. To find the resistance, we rearrange the formula:
$$R = \frac{V}{I}$$

**Calculation:**
$$R = \frac{12\ \text{V}}{0.5\ \text{A}}$$
$$R = 24\ \Omega$$

The resistance of the resistor is **$24\ \Omega$**.

---

### (b) Electrical Power Dissipated ($P$)

The electrical power dissipated can be calculated using the formula:
$$P = V \times I$$
(Alternatively, $P = I^2R$ or $P = \frac{V^2}{R}$)

Using the voltage and current values (the most direct method):

**Calculation:**
$$P = 12\ \text{V} \times 0.5\ \text{A}$$
$$P = 6\ \text{W}$$

The electrical power dissipated is **$6\ \text{W}$**.

---

### Summary of Results:
(a) Resistance ($R$) = $\mathbf{24\ \Omega}$
(b) Power ($P$) = $\mathbf{6\ \text{W}}$

To solve the problem, we'll use **Ohm's Law** and the formula for electrical power.

### (a) Calculating the Resistance of the Resistor

**Ohm's Law** states:
\[
V = I \times R
\]
Where:
- \( V \) is the voltage across the resistor,
- \( I \) is the current through the resistor,
- \( R \) is the resistance.

We are given:
- \( V = 12 \, \text{V} \)
- \( I = 0.5 \, \text{A} \)

Rearranging the formula to solve for \( R \):
\[
R = \frac{V}{I}
\]
Substituting the given values:
\[
R = \frac{12 \, \text{V}}{0.5 \, \text{A}} = 24 \, \Omega
\]

**Final Answer:**
\[
\boxed{24 \, \Omega}
\]

### (b) Calculating the Electrical Power Dissipated

The **power dissipated** in a resistor can be calculated using:
\[
P = V \times I
\]
Where:
- \( P \) is the power,
- \( V \) is the voltage,
- \( I \) is the current.

Using the given values:
\[
P = 12 \, \text{V} \times 0.5 \, \text{A} = 6 \, \text{W}
\]

**Final Answer:**
\[
\boxed{6 \, \text{W}}
\]

To solve this problem, we'll use Ohm's Law: 

V = IR 
 
Where: 
- V is the Voltage across the circuit or component. 
- I is the Current flowing through the circuit or component. 
- R is the Resistance of the resistor in the circuit. 

(a) The resistance of the resistor (R): 

We know that the voltage across the battery is $12\text{ V}$ and the current flow is $0.5\text{ A}$. We'll use these to calculate: 

$$ 
\begin{align*} 
     R &= \frac {V}{I} \\ 
     &= \frac{12\text{ V}}{0.5\text{ A}} \\ 
     &=  24 \,\Omega . 
 \end{align*}  
 $$ 

So, the resistance of the resistor is $24$ Ohms. 

(b) The power dissipated in the resistor (P): 

Now that we know the voltage across and current through the circuit, and have calculated the resistance above, we can calculate the electrical power using: 

$$ 
\begin{align*} 
     P &= IV \\ 
     &= I^2R \\
     &= \frac {V^2}{R} .  
 \end{align*}  
 $$ 

Using these three formulas to solve for $P$ gives us: 

$$ 
\begin{align*} 
     P_1 &= (0.5\text{ A})(12\text{ V}) \\
     & =  6 \,\text{W} . 
 \end{align*}  
 $$ 
 
and, 
 
$$ 
\begin{align*} 
     P_2 &= (0.5\text{ A})^2 (24\Omega) \\
     &= 0.375 \,\text{A}^2/\Omega * 24\Omega\\
     & =  9 \,\text{W} . 
 \end{align*}  
 $$ 
 
and, 
 
$$ 
\begin{align*} 
     P_2 &= \frac {(12\text{ V})^2}{24\Omega} \\
     &= \frac {144\text{ V}^2/\Omega}{24\Omega}\\ 
     & =  6 \,\text{W} . 
 \end{align*}  
 $$ 
 
Therefore, the electrical power dissipated in the resistor is $6$ Watts.

(a) Using Ohm's Law \( V = IR \), we solve for resistance \( R \):

\[
R = \frac{V}{I} = \frac{12\ \text{V}}{0.5\ \text{A}} = 24\ \Omega
\]

**Answer:** The resistance of the resistor is \(\boxed{24\ \Omega}\).

(b) To find the electrical power dissipated, we use \( P = VI \):

\[
P = V \times I = 12\ \text{V} \times 0.5\ \text{A} = 6\ \text{W}
\]

**Answer:** The electrical power dissipated is \(\boxed{6\ \text{W}}\).