In a parallel circuit, there is only one path for current to take. Please select the best answer from the choices provided T F

1 answer:

7 0

PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG. PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG.

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If I have a picosecond laser how would that be expressed in terms of zeptoseconds? In terms of petaseconds?

kolbaska11 [484]

Picosecond = 10 ^ -12 seconds.
Zeptosecond = 10^ -18 seconsds
Petaseonds = 10^15 seconds

To express Picoseconds into any of other two, you have to divide 10^-12 by the power index of the one in question

1Picosecond : 10^-12 / 10^-18 = 10^ (-12- 18) = 10^ (-12+18)= 10^6 zeptoseconds

1Picosecond : 10^-12 / 10^15 = 10^ (-12-15) = 10^-27 Petaseconds.

1Picosecond = 10^6 zeptoseconds

1Picosecond = 10^-27 Petaseconds

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3 years ago

Can someone help with me 1,2,3 please I will mark brainless .

Yuki888 [10]

Answer:

1) A. .33 hr

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2 years ago

Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an

ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

$Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega$