If you are in a car at a stop sign and then the driver

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

insens350 [35]

(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.

7 0

3 years ago

What is terrestrial radiation?

Feliz [49]

C) Radiation that comes from Earth...... Hope it helps, Have a nice day :)

6 0

3 years ago

Read 2 more answers

A 400g sample of water absorbs 500j of energy. how did the water temperature change if the specific heat of water is 4.18j/g©. S

Mashutka [201]

In general, the quantity of heat energy, Q, required to raise a mass m kg of a substance with a specific heat capacity of c J/(kg °C), from temperature t1 °C to t2 °C is given by:

Q = mc(t2 – t1) joules

So:

(t2-t1) =Q / mc

As we know:
Q = 500 J

m = 0.4 kg

c = 4180 J/Kg °c

We can take t1 to be 0°c

t2 - 0 = 500 / ( 0.4 * 4180 )

t2 - 0 = 0.30°c

6 0

3 years ago

Multiply 5.036×102m by 0.078×10−1, taking into account significant figures.

Anika [276]

Answer : The significant digit is 6

Explanation :

Multiply $5.036\times10^{2}m$ by $0.078\times10^{-1}m$

Now, on multiplying

$5.036\times10^{2}m \times0.078\times10^{-1}m = 0.392808 \times10^{1}\ m^{2}$

$5.036\times10^{2}m \times0.078\times10^{-1}m = 0.0392808\ m^{2}$

Now, the significant digit is 6.

Hence, this is the required solution.

3 0

3 years ago

What is the Validity measures of a basketball bouncing

kykrilka [37]

This study was aimed at testing the construct validity of the basketball basic motion skills test instrument (ITK GDBB). The research used descriptive method of 3 basketball experts in the city of Cimahi; 3 experts are the expert in basketball. The instrument used was the ITB GDBB developed by Silvy (2019) consisting of top passing, bottom passing, top service, bottom service, chest passing, bounding passing, overhead passing, and leading ball (dribbling). This instrument consists of 76 items that cover 4 domains in basketball, namely chest pass, overhead pass, bound pass, and dribbling. The validity method used the construct validity of different power types. For the reliability method, it used the Kuder Ricardson (KR) and Objectivity analysis. The results of the construct validity analysis of a total of 76 items show that the score is ranged from 0.67 to 1.00. The construct validity value of 71 items in the basketball game is in the high category (= 1.00), 5 items are in the sufficient category, the relativity score is ranged from 0.75 to 0.98, and the objectivity score is ranged from 0.89 to 0.95. The conclusion is that this test instrument can be used as a standardized basic motion skill test for standardized large ball games for validity in basic motion skills in basketball games for grade VII junior high school students.

3 0

3 years ago