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3 years ago

14

An electric heater is rated at 1400 W, a toaster is rated at 1150 W, and an electric grill is rated at 1560 W. The three applian

ces are connected in parallel across a 112 V emf source. Find the current in the heater.

1 answer:

gayaneshka [121]3 years ago

8 0

Answer:

The current in the heater is 12.5 A

Explanation:

It is given that,

Power of electric heater, P₁ = 1400 W

Power of toaster, P₂ = 1150 W

Power of electric grill, P₃ = 1560 W

All three appliances are connected in parallel across a 112 V emf source. We need to find the current in the heater. We know that in parallel combination of resistors the current flowing in every branch of resistor divides while the voltage is same.

Electric power, $P_1=V\times I_1$

$I_1=\dfrac{P_1}{V}$

$I_1=\dfrac{1400\ W}{112\ V}$

$I_1=12.5\ A$

So, the current in the heater is 12.5 A. Hence, this is the required solution.

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Answer:

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Explanation:

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1 year ago

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3 years ago

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

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Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

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Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

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