Azzalictpdv
2022-05-19
Answered

What is the electric current produced when a voltage of 24V is applied to a circuit with a resistance of $8\mathrm{\Omega}$?

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stormiinazlhdd

Answered 2022-05-20
Author has **12** answers

Explanation:

It is 3 A because as the Ohm's law formula is V=IR

where, V is the voltage

I is the electric current

and R is the resistance.

So by taking the values we get,

V=IR

$24V=I\times 8\mathrm{\Omega}$

$I=24\xf78$

I=3A

It is 3 A because as the Ohm's law formula is V=IR

where, V is the voltage

I is the electric current

and R is the resistance.

So by taking the values we get,

V=IR

$24V=I\times 8\mathrm{\Omega}$

$I=24\xf78$

I=3A

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This is kind of a stupid question and I am taking some risk of getting some down-votes here, but, I can't resist posting it. Suppose $({u}_{1},{u}_{2})$ is an orthonormal basis for ${R}^{2}$, and let x be an arbitrary vector in ${R}^{2}$, then we can decompose x by projecting on ${u}_{1},{u}_{2}$, i.e.

$\begin{array}{r}({u}_{1}^{T}x){u}_{1}\\ ({u}_{2}^{T}x){u}_{2}\end{array}$

Certainly we have

$\begin{array}{r}x=({u}_{1}^{T}x){u}_{1}+({u}_{2}^{T}x){u}_{2}\end{array}$

Looks like a simple problem, but how can I prove this?

$\begin{array}{r}({u}_{1}^{T}x){u}_{1}\\ ({u}_{2}^{T}x){u}_{2}\end{array}$

Certainly we have

$\begin{array}{r}x=({u}_{1}^{T}x){u}_{1}+({u}_{2}^{T}x){u}_{2}\end{array}$

Looks like a simple problem, but how can I prove this?

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