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3 years ago

Which circuit is a series circuit ?? plz help

Physics

2 answers:

gregori [183]3 years ago

7 0

In a series circuit, all of the components are connected in the same 'loop' and the current only has one direction/path it can flow through.

In the first three options, the current has multiple paths it can go through. So these three circuits are parallel and not series.

In the last option, the current only has one path where it can flow through, so that circuit is in series.

So Circuit <u>D </u>is a series circuit.

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Answer

Circuit D

Ugo [173]3 years ago

5 0

Answer:the last one

Explanation:in a series circuit, only one current flows through it. i.e it connected by a single wire

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1 year ago

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3 years ago

Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at

sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

E = k*Q / r^2

Where, k: Coulomb's Constant

Q: The charge of particle

r: The distance from source

- The Electric Field due to charge 1:

E_1 = k*Q_1 / r^2

E_1 = k*Q / (2*a)^2

E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

E_2 = k*Q_2 / r^2

E_2 = k*Q_2 / (13*a)^2

E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

E_net = E_1 + E_2

2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

1: 2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

2Q = Q / 4 + Q_2 / 169

Q_2 = 295.75*Q

2: 2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

2Q = Q / 4 - Q_2 / 169

Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0

3 years ago

Look at the circuit below. What is the voltage between points C and D?

stepan [7]

Option c) 1.5 V

Explanation:

<em>As the circuit is build in series first we will find the current passing through the complete circuit. Current stays the same in each element is the series cirucuit, however, the voltage is different.</em>

Voltage is given by the following formula:

V = IR

<em>Because we have to find current through whole circuit, we will first find resistance of the whole circuit.</em>

Equivalent Resistance R(eq): R1 + R2 = 60 + 60 = 120 ohm

Current passing through whole circuit be:

$I = \frac{V}{R(eq)}\\I = \frac{3}{120}$ = 0.025

Now we will find out the voltage between C and D:

Current stays the same in series circuit: I = 0.025 c

Resistance between C and D is, R = 60 ohm

Voltage becomes, V = IR = 0.025 * 60 = 1.5 V

7 0

3 years ago

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