All categories

3 years ago

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

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

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

Peak current through resistor is independent of frequency

I₂ = 2.0 A

You might be interested in

A spring (k = 200 N/m) is fixed at the top of a frictionless plane inclined at angle θ = 33 °. A 1.2 kg block is projected up th

Serga [27]

Answer:

Explanation:

Initial kinetic energy = 29 J

work done against gravity = mgsin33 x d , m is mass of the block

= 1.2 x 9.8 sin 33 x .9

= 5.76 J

potential energy stored in compressed spring

= 1/2 k x², k is spring constant and x is compression

= .5 x 200 x .3²

= 9

energy left = 29 - ( 5.76 + 9 )

= 14.24 J

b )

energy stored in spring when compression is .4 m

= 1/2 x 200 x .4²

= 16 J

required kinetic energy = 16 + 5.76

= 21.76 J

Block must be projected with energy of 21.76 J .

7 0

3 years ago

A proton moves from left to right across the screen at speed v0. Then a uniform magnetic field of strength 0.80 T is directed in

Anton [14]

Answer: The force is directed upward

Explanation: Considering the Lorentz force, given by:

F= qv×B

using the right hand rule and considering the direction of electron velocity and the magnetic field from the figure, the vectorial product gives a force vector upwards .

8 0

4 years ago

True or False. The two metals in a battery are the electrolytes. True False

liubo4ka [24]

Falseeeeeeeeeeeeeeeeeeeeeeeeeeeee

6 0

3 years ago

Read 2 more answers

The pressure exerted by a gas is 2.0 atm while it has a volume of 350 mL. What would be the volume of this sample of gas at stan

34kurt

Answer:

700 mL or 0.0007 m³

Explanation:

P₁ = Initial pressure = 2 atm

V₁ = Initial volume = 350 mL

P₂ = Final pressure = 1 atm

V₂ = Final volume

Here the temperature remains constant. So, Boyle's law can be applied here.

P₁V₁ = P₂V₂

$\frac{P_1V_1}{P_2}=V_2\\\Rightarrow V_2=\frac{2\times 350}{1}\\\Rightarrow V_2=700\ mL$

So, volume of this sample of gas at standard atmospheric pressure would be 700 mL or 0.0007 m³

7 0

4 years ago

A cylinder with rotational inertia I1=2.0kg·m2 rotates clockwise about a vertical axis through its center with angular speed ω1=

Mrac [35]

Answer:

a) 0.67 rad/sec in the clockwise direction.

b) 98.8% of the kinetic energy is lost.

Explanation:

Let us take clockwise angular speed as +ve

For first cylinder

rotational inertia $I$ = 2.0 kg-m^2

angular speed ω = +5.0 rad/s

For second cylinder

rotational inertia $I$ = 1.0 kg-m^2

angular speed = -8.0 rad/s

The rotational momentum of a rotating body is given as = $I$ ω

where $I$ is the rotational inertia

ω is the angular speed

The rotational momenta of the cylinders are:

for first cylinder = $I$ ω = 2.0 x 5.0 = 10 kg-m^2 rad/s

for second cylinder = $I$ ω = 1.0 x (-8.0) = -8 kg-m^2 rad/s

The total initial angular momentum of this system cylinders before they were coupled together = 10 + (-8) = 2 kg-m^2 rad/s

When they are coupled coupled together, their total rotational inertia $I_{t}$ = 1.0 + 2.0 = 3 kg-m^2

Their final angular rotational momentum after coupling = $I_{t}$ $w_{f}$

where $I_{t}$ is their total rotational inertia

$w_{f}$ = their final angular speed together

Final angular momentum = 3 x $w_{f}$ = 3 $w_{f}$

According to the conservation of angular momentum, the initial rotational momentum must be equal to the final rotational momentum

this means that

2 = 3 $w_{f}$

$w_{f}$ = final total angular speed of the coupled cylinders = 2/3 = 0.67 rad/s

From the first statement, the direction is clockwise

b) Rotational kinetic energy = $\frac{1}{2} Iw^{2}$

where $I$ is the rotational inertia

$w$ is the angular speed

The kinetic energy of the cylinders are:

for first cylinder = $\frac{1}{2} Iw^{2}$ = $\frac{1}{2}*2*5^{2}$ = 25 J

for second cylinder = $\frac{1}{2}*1*8^{2}$ = 32 J

Total initial energy of the system = 25 + 32 = 57 J

The final kinetic energy of the cylinders after coupling = $\frac{1}{2}I_{t}w^{2} _{f}$

where

where $I_{t}$ is the total rotational inertia of the cylinders

$w_{f}$ is final total angular speed of the coupled cylinders

Final kinetic energy = $\frac{1}{2}*3*0.67^{2}$ = 0.67 J

kinetic energy lost = 57 - 0.67 = 56.33 J

percentage = 56.33/57 x 100% = 98.8%

6 0

3 years ago