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2 years ago

What is the value of the equivalent resistance for the three resistors connected in series?

1 answer:

7 0

The value of the equivalent resistance for the three resistors connected in series will be the sum of the three values.

To find the answer, we have to know more about the equivalent resistance.

<h3>What is meant by equivalent resistance?</h3>

equivalent resistance is the total value of the resistance connected in a circuit.
If n resistors are connected in series, then the equivalent resistance will be,

$R_E=R_1+R_2+..........+R_n$

In our question we have three resistors. Thus, the equivalent resistance will be,

$R_E=R_1+R_2+R_3$

Thus, we can conclude that, the value of the equivalent resistance for the three resistors connected in series will be the sum of the three values.

Learn more about the equivalent resistance here:

brainly.com/question/11603204

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Can someone solve this problem and explain to me how you got it

likoan [24]

Answer:

Explanation:

Coulomb's law states that the force of attraction or repulsion between any two charges is proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between the charges

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question 7:

Given: $F=1.9*10^{-29}N$ , $q1=q2=1.6*10^{-19}C$

By coulomb's law,

$1.9*10^{-29}=9*10^{9}*\frac{1.6*10^{-19}*1.6*10^{-19}}{r^{2}}$

⇒ $r=3.5m$

Question 6:

Given: $q1=q2=-1.5*10^{-6}C$ , $r=0.28m$

By coulomb's law,

$F=9*10^{9}*\frac{1.5*10^{-6}*1.5*10^{-6}}{0.28^{2}}$

⇒ $F=0.26N$

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