In a string of lights, if one light goes out and the others stay lit, which type of circuit do you have?

Physics

2 answers:

Svetradugi [14.3K]4 years ago

8 0

Answer:

A parallel circuit

Explanation:

The light bulbs can be connected in two different configurations:

1. Series --> when connected in series, the light bulbs are on the same branch of the circuit. This means that the same current flows through all the light bulbs, while the potential difference across each light can be different.

2. Parallel --> when connected in parallel, the light bulbs are on separate branches of the circuit. This means that the current flowing through each bulb is different, while they all have same potential difference across them.

In the series configuration, when one light goes out, all the other lights also turn off, because they are all on the same branch: so if one breaks, it opens the circuit, and the current can no longer flow through the other lights.

On the other hand, in a parallel configuration, when one light goes out, the other lights stay on, because they are on separate branches, so they are not affected by the broken branch and current can still flow through them.

zepelin [54]4 years ago

3 0

You have a parallel circuit since it allows the electrons to travel in 2 paths, allowing the other light to stay lit.

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If position vector r = bt^2i + ct^3j, where b and c are positive constants, when does the velocity vector make an angle of 450 w

vazorg [7]

$\vec r(t)=bt^2\,\vec\imath+ct^3\,\vec\jmath$

The velocity at time $t$ is

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=2bt\,\vec\imath+3ct^2\,\vec\jmath$

Take two vectors that point in the positive $x$ and positive $y$ directions, such as $\vec\imath$ and $\vec\jmath$ . The dot products of the velocity vector with $\vec\imath$ and $\vec\jmath$ are

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\imath=2bt=\sqrt{4b^2t^2+9c^2t^4}\cos\theta$

and

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\jmath=3ct^2=\sqrt{4b^2t^2+9c^2t^4}\cos\theta$

We want the angles between these vectors to be 45º, for which we have $\cos45^\circ=\frac1{\sqrt2}$ . So

$\begin{cases}2\sqrt2\,bt=\sqrt{4b^2t^2+9c^2t^4}\\3\sqrt2\,ct^2=\sqrt{4b^2t^2+9c^2t^4}\end{cases}\implies3\sqrt2\,ct^2-2\sqrt2\,bt=0$

$\implies t(3ct-2b)=0$

$\implies t=0\text{ or }t=\dfrac{2b}{3c}$

When $t=0$ , the velocity vector is equal to the zero vector, which technically has no direction/doesn't make an angle with any other vector. So the only time this happens is for