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aev [14]
3 years ago
10

At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does

the potential difference across the resistor compare to the emf across the coil? (Enter your answers in V.) resistor V coil V (b) Answer the same question about the circuit several seconds later. (Enter your answers in V.) resistor V coil V (c) Is there an instant at which these two voltages are equal in magnitude? Yes No (d) If so, when? Is there more than one such instant? (Enter all possible times in ms as a comma-separated list. If there are no such instants, enter NONE.) ms (e) After a 3.20 A current is established in the resistor and coil, the battery is suddenly replaced by a short circuit. Answer questions (a) and (b) again with reference to this new circuit. (Enter your answers in V.) immediately thereafter several seconds later resistor V V Need Help?
Physics
1 answer:
insens350 [35]3 years ago
7 0

(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

V = V_R + V_L

The potential difference across the inductance is given by

V_L(t) = V e^{-\frac{t}{\tau}} (1)

where

\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s is the time constant of the circuit

At time t=0,

V_L(0) = V e^0 = V = 20 V

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

V_R = V-V_L = 20 V-20 V=0

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

t >> \tau

Since \tau is at the order of less than milliseconds.

Using eq.(1), we see that for t >> \tau, the exponential becomes zero, and therefore the potential difference across the coil is zero:

V_L = 0

Therefore, the potential difference across the resistor will be

V_R = V-V_L = 20 V- 0 = 20 V

(c) Yes

The two voltages will be equal when:

V_L = V_R (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

V=V_L +V_R

And rewriting this equation,

V_R = V-V_L

Substituting into (2) we find

V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V

So, the two voltages will be equal when they are both equal to 10 V.

(d) at t=5.77\cdot 10^{-4}s

We said that the two voltages will be equal when

V_L=\frac{V}{2}

Using eq.(1), and this last equation, this means

V e^{-\frac{t}{\tau}} = \frac{V}{2}

And solving the equation for t, we find the time t at which the two voltages are equal:

e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s

(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

I_0 = 3.20 A

The problem says this current is stable: this means that we are in a situation in which t>>\tau, so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

V_L(t) = -V e^{-\frac{t}{\tau}} (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

V_L(0)=.V

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

t>>\tau

If we use this approximation into the formula

V_L(t) = -V e^{-\frac{t}{\tau}} (1)

We find that

V_L = 0

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

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From the experiment, the constructed conclusion was that;

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When energy from a battery is added to water, two streams of gas bubbles is said to be  produced. Then a lit match is put into  trapped gas from these two sources, one pops and the other glows brighter.

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A delivery man pushes the button to ring the doorbell. Which of the following forms of energy is not involved in this process
mihalych1998 [28]

Of the following given choices;

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The answer is; gravitational.

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3 0
3 years ago
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A ceramic tile measuring 50 cm x50cm has been designed to bear a pressure of 40 N/in . Will it with stand a force of 5 N?
dangina [55]

Answer:

The ceramic tile will stand the force of 5 N.

Explanation:

Step 1: Calculate the area (A) of the ceramic tile

We will use the formula for the area of a square.

A = 50 cm × 50 cm = 2500 cm²

Step 2: Convert "A" to in²

We will use the conversion factor 1 in² = 6.45 cm².

2500 cm² × 1 in²/6.45 cm² = 3.9 × 10² in²

Step 3: Calculate the pressure (P) exerted by a force (F) of 5 N

We will use the following expression.

P = F/A

P = 5 N / 3.9 × 10² in² = 0.013 N/in²

Since the pressure exerted would be less than the maximum pressure resisted (40 N/in²), the ceramic tile will stand the force of 5 N.

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2 years ago
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5 0
3 years ago
A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on
g100num [7]

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J  

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J  

6 0
3 years ago
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