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murzikaleks [220]
3 years ago
15

In a series RL circuit, ET = 120 V, R = 40 Ω, and XL = 30 Ω. What is ER?

Mathematics
1 answer:
Delvig [45]3 years ago
7 0

Answer:

The voltage across resistor is E_{R}=96V

Step-by-step explanation:

We are given with Voltage across terminals of a series RL circuit, E_{T}=120V

And Resistance R=40Ω

And inductor reactance X_{L}=30Ω

Since Resistor and inductor are in series, same current flows through both.

And that current, I = \frac{E_{T}} {Z}

                             =\frac{120}{\sqrt{40^{2}+30^{2}}}= \frac{120}{\sqrt{2500}}

                            = \frac{120}{50}=2.4A

Therefore voltage across resistor, E_{R}=IR=2.4X40=96V

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3 years ago
Which of the following matrices is the solution matrix for the given system of equations? x + 5y = 11 x - y = 5
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<h2>The required solution is x = 6 and y = 11 </h2>

Step-by-step explanation:

Given system of equations are

x+5y = 11 and x-y =5

A= \left[\begin{array}{cc}1&5\\1&-1\end{array}\right]                            X=\left[\begin{array}{c}x\\y\end{array}\right]

and          B= \left[\begin{array}{c}11\\5\end{array}\right]

∴AX=B

adj A = \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]

A= \left|\begin{array}{cc}1&5\\1&-1\end{array}\right|=-6

∴A^{-1} =\frac{adj A}{|A|}

So,A^{-1} =\frac{ \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]}{-6}

A^{-1} ={ \left[\begin{array}{c \c}  {{\frac{1}{6} }}&{\frac{5}{6}}\ \\  {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]}

X =A^{-1}\times B

⇒\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c \c}  {{\frac{1}{6} }}&{\frac{5}{6}}\ \\  {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]} \times \left[\begin{array}{c}11\\5\end{array}\right]

⇒\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c}  {6}\\  {11} \end{array}\right]}

∴ x= 6 and y = 11

The required solution is x = 6 and y = 11

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