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DerKrebs [107]
3 years ago
5

A resistor is connected across an oscillating emf. The peak current through the resistor is 2.0 A. What is the peak current if:

Physics
1 answer:
Salsk061 [2.6K]3 years ago
6 0

Answer:

(a) When the resistance R is doubled, I = 1 A

(b) When the peak emf εo is doubled, I = 4 A

(c)  When the frequency ω is doubled, I = 2 A

Explanation:

Given;

peak current through the resistor, I = 2.0 A

According to ohms law the peak current through the circuit is given by;

I = \frac{V}{R}

(a) When the resistance R is doubled;

I = \frac{V_R}{R} \\\\I_1R_1 = I_2R_2\\\\I_2 = \frac{I_1R_1}{R_2} \\\\I_2 =  \frac{2*R_1}{2R_1} \\\\I_2 = 1 \ A

(b)When the peak emf εo is doubled

I = \frac{V}{R} = \frac{\epsilon_o}{R} \\\\R = \frac{\epsilon_ o}{I} \\\\\frac{\epsilon_ o_1}{I_1}  = \frac{\epsilon_ o_2}{I_2} \\\\I_2 = \frac{\epsilon_ o_2 *I_1}{\epsilon _o_1} \\\\I_2 = \frac{2 \epsilon_ o_1 *2}{\epsilon _o_1} \\\\I_2 = 4 \ A

(c) When the frequency ω is doubled

Peak current through resistor is independent of frequency

I₂ = 2.0 A

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Given \sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] at a point. What is the force per unit area at this point acting normal to the surface with\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z   ? Are there any shear stresses acting on this surface?

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