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balandron [24]
3 years ago
8

A circuit element maintains a constant resistance. If the current through the circuit element is doubled, what is the effect on

the power dissipated by the circuit element
Physics
1 answer:
Morgarella [4.7K]3 years ago
3 0

Answer:

<h3>This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.</h3>

Explanation:

The formula for calculating the power expended in a circuit is P =  I²R... 1

i is the current (in amperes)

R is the resistance (in ohms)

If  a circuit element maintains a constant resistance and the current through the circuit element is doubled, then new current I₂ = 2I

New power dissipated P₂ = (I₂)²R

P₂ = (2I)²R

P₂ = 4I²R ... 2

Dividing equation 2 by 1 will give;

P₂/P = 4I²R/I²R

P₂/P = 4

P₂ = 4P

This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.

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A mass hanging from a spring oscillates with a period of 0.35 s. Suppose the mass and spring are swung in a horizontal circle, w
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Answer:

66 rpm

Explanation:

The period of oscillation is given by

T=2\pi \sqrt{\frac {m}{k}}

\frac {k}{m}=\frac {4\pi^{2}}{T^{2}} where  T is time period of oscillation which is given as 0.35 s, k s spring constant and m is the mass of the object attached to the spring.

Also, net force is given by

Net force=m\omega^{2} L

\omega=\sqrt{\frac {k\triangle L}{mL}} where \triangle L is the elongation, L is original length, \omega is the angular velocity

Substituting the equation of \frac {k}{m} into the above we obtain

\omega=\sqrt {\frac {4\pi^{2}\triangle L}{T^{2} L}}

\omega=\sqrt {4\pi^{2}\times 0.15L}{0.35^{2}\times L}}=6.952763\approx 6.95 rad/s

6.95\times\frac {60 s}{2\pi rad}\approx 66 rpm

6 0
3 years ago
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An ideal transformer has 60 turns in its primary coil and 360 turns in its secondary coil. If the input rms voltage for the 60-t
notka56 [123]

Answer:

720 V

Explanation:

Given that,

The number of turns in primary coil, N₁ = 60

The number of turns in secondary coil, N₂ = 360

The input rms voltage, V₁ = 120 V

We need to find the output rms voltage of the secondary coil . The relation between number of turns in primary coil - secondary coil to the input rms voltage to the output rms voltage is given by :

\dfrac{N_1}{N_2}=\dfrac{V_1}{V_2}\\\\V_2=\dfrac{N_2V_1}{N_2}\\\\V_2=\dfrac{360\times 120}{60}\\\\V_2=720\ V

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