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slega [8]
2 years ago
6

A circuit is built based on this circuit diagram. What is the equivalent resistance of the circuit?

Physics
1 answer:
Dmitry_Shevchenko [17]2 years ago
7 0
The three resistors are connected to the same points of the circuit, so they are in parallel configuration. The equivalent resistance of 3 resistors in parallel is given by:
\frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}
If we plug the values of the resistances into the formula, we find
\frac{1}{R_{eq}}= \frac{1}{3.0 \Omega}+ \frac{1}{6.0 \Omega}+ \frac{1}{9.0 \Omega}= \frac{6+3+2}{18.0 \Omega} = \frac{11}{18.0 \Omega}
From which we find the equivalent resistance:
R_{eq} =  \frac{18.0}{11} \Omega =1.6 \Omega

So, the correct answer is B.
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The net electric field at point D is determined as 3.95 x 10⁷ N/C.

<h3>Electric field at D due to charge A</h3>

E = kq/r²

where;

  • r is the distance between A and D
  • q is charge A

E(AD) = (9 x 10⁹ x 6 x 10⁻⁶)/(0.05²)

E(AD) = 2.16 x 10⁷ i N/C

<h3>Electric field at D due to charge B</h3>

E = kq/r²

where;

  • r is the distance between A and B
  • q is charge B

r² = 5²  + 5²

r² = 50

r = √50

r = 7.07 cm

E(BD) = (9 x 10⁹ x 5 x 10⁻⁶)/(0.0707²)

E(BD) = 9 x 10⁶ N/C

in x - direction = 9 x 10⁶ N/C x cos(45) = 6.36 x 10⁶ i  N/C

in y - direction = 9 x 10⁶ N/C x sin(45) = 6.36 x 10⁶j  N/C

<h3>Electric field at D due to charge C</h3>

E = kq/r²

where;

  • r is the distance between C and D
  • q is charge C

E(CD) = (9 x 10⁹ x 6 x 10⁻⁶)/(0.05²)

E(CD) = 2.16 x 10⁷ j N/C

<h3>Net electric field in x direction</h3>

Ei = 2.16 x 10⁷ i N/C  + 6.36 x 10⁶ i  N/C

Ei = 2.796 x 10⁷ i N/C

<h3>Net electric field in y direction</h3>

Ej = 2.16 x 10⁷ j N/C  + 6.36 x 10⁶j  N/C

Ej = 2.796 x 10⁷ j N/C

<h3>Resultant electric field at D</h3>

E = √Ei² + Ej²

E = √[(2.796 x 10⁷)² + (2.796 x 10⁷)²]

E = 3.95 x 10⁷ N/C

Thus, the net electric field at point D is determined as 3.95 x 10⁷ N/C.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

7 0
1 year ago
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