Series circuit containing 2 light bulbs, three wires, and a power source. What would happen if one of the light bulbs burnt out? Why would this happen?

The correct answer is no bulb will glow if one burns out because the current flow is disrupted.

Explanation:

We make a series connection of two bulbs with the help of 3 wires and a power source.

One bulb burns out.

Therefore the circuit is now broken and the current cannot flow from one end to the other end of the circuit as because the bulb that burnt out used to help in completing the circuit.

Thus no bulb lights up when current is passed in the connection.

answer: the figure is attached herewith.

step-by-step explanation:   we are given to sketch a unit circle and label the terminal points corresponding to

in the attached figure, we have drawn a unit circle, with centre at o(0,0) and radius 1 unit. we have labelled the points a, b, c, d and e corresponding to the angles below

∠aox = 60°,

∠box = 120°,

∠cox = 180°,

∠dox = 240°,

∠eox = 300°.

and the coordinates are a(0.5,1), b(-0.5,1), c(-1,0), d(-0.5,-1) and e(0.5,-1).

thus, the answer is complete.