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3 years ago

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The RMS potential difference of an AC household outlet is 117 V. a) What is the maximum potential difference across a lamp conne

cted to the outlet? b) If the RMS current through the lamp is 5.5 A, what is the lam

1 answer:

Anit [1.1K]3 years ago

6 0

Answer:

a. 165.5 V

b. 7.78 A

Explanation:

Here is the complete question

The RMS potential difference of an AC household outlet is 117 V. a) What is the maximum potential difference across a lamp connected to the outlet? b) If the RMS current through the lamp is 5.5 A, what is the maximun current through the lamp.

Solution

a. The maximum potential difference across the lamp V₀ = √2V₁ where V₁ = rms value of potential difference = 117 V

V₀ = √2V₁ = √2 × 117 V = 165.5 V

b. The maximum current through the lamp I₀ = √2I₁ where I₁ = rms value of current = 5.5 A

V₀ = √2V₁ = √2 × 5.5 A = 7.78 A

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A point charge is at the origin. With this point charge as the source point, what is the unit vector r^ in the direction of (a)

KiRa [710]

Answer:

a. $\hat{r} =- \hat{j}$
b. $\hat{r} = \ \frac{1}{\sqrt{2}} \ \hat{i} + \ \frac{1}{\sqrt{2}} \ \hat{j}$
c. $\hat{r} = \ -0.3871 \ \hat{i} + \ 0.91501\ \hat{j}$

Explanation:

Using Coulomb's Law we know that the electric field E at point $\vec{r}$ is:

$\vec{E(\vec{r})} = k_e \frac{q}{d^2} \frac{\vec{r}-\vec{r'}}{d}$

where $k_e$ is the Coulomb's Constant, q is the source charge, d is the distance between point and position of the source point charge, and $\vec{r}'$ is the position of the source point charge.

Taking all this in consideration, the unit vector clearly is:

$\hat{r} =\frac{\vec{r}-\vec{r'}}{d}$

For our problem, $\vec{r'} = (0,0)$ , as the charge is located at the origin.

So

$\hat{r} =\frac{\vec{r}}{d}$

and d will be the magnitude of $\vec{r}$

Now, we can take the values for each point.

<h3>a.</h3>

$\vec{r}= (0,-1.35 \ m)$

and, the magnitude of the vector is

$|\vec{r}| = \sqrt{r_x^2 + r_y^2}$

$|\vec{r}| = \sqrt{(0 \ m)^2 + (-1.35 \ m )^2}$

$|\vec{r}| =1.35 \ m$

So, the unit vector is:

$\hat{r} =\frac{(0,-1.35 \ m)}{1.35 \ m}$

$\hat{r} =(0,-1,0)$

$\hat{r} =- \hat{j}$

<h3>b.</h3>

$\vec{r}= (12 \ cm,12 \ cm)$

and, the magnitude of the vector is

$|\vec{r}| = \sqrt{r_x^2 + r_y^2}$

$|\vec{r}| = \sqrt{(12 \ cm)^2 + (12 \ cm )^2}$

$|\vec{r}| = \sqrt{2} \ 12 \ cm$

So, the unit vector is:

$\hat{r} =\frac{(12 \ cm,12 \ cm)}{\sqrt{2} \ 12 \ cm}$

$\hat{r} =(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$

$\hat{r} = \ \frac{1}{\sqrt{2}} \ \hat{i} + \ \frac{1}{\sqrt{2}} \ \hat{j}$

<h3>c.</h3>

$\vec{r}= (-1.10 \ m, 2.60 \ m)$

and, the magnitude of the vector is

$|\vec{r}| = \sqrt{r_x^2 + r_y^2}$

$|\vec{r}| = \sqrt{(-1.10 \ m)^2 + (2.60 \ m)^2}$

$|\vec{r}| = 2.8415 \ m$

So, the unit vector is:

$\hat{r} =\frac{ (-1.10 \ m, 2.60 \ m)}{2.8415 \ m}$

$\hat{r} =(-0.3871 ,0.91501)$

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