Sophomore Physics Elective 3 1 Chapter 2 Section 7 Problem Solving

1) By "When the switch is closed, the total power of the power supply is 16 watts, and the output power is 12 watts, at which point the bulb glows normally. "Yes.

Total e*i total 16 watts.

p out u*i total 12 watts.

pin p total p out i total 2* r 16 12 4 watts.

So the total current is i total 2 amps (due to the internal resistance of the power supply r 1 ohm), the electromotive force e 8 volts, and the road terminal voltage u 6 volts.

The current in R2 is i2 u R2 6 6=1 amps.

The current in R1 is i1 i, total i2, 1 1 amps.

Since the current of the two branches is equal, there is r2 r1 r lamp.

The resistance of the lamp is r2 r1 6 2 4 ohms.

2) When the switch is disconnected, the bulb can still emit light normally, that is, to ensure that the current of the lamp is still 1 ampere.

Let the value of r1 be r1x

Then i light e (r1x r light r).

1＝8 / (r1x＋4＋1 )

The resistance value of the rheostat is r1x 3 ohms.

At this time, the road-end voltage is u e i lamp*r 8 1*1 7 volts.

The power output of the power supply is p out i lamp*u 1*7 7 watts.

The power efficiency is 7 8*100%.

Solution: (1) The power consumed by the internal resistance r.

p in = p total - p out = 16-12 = 4w

p inside = i r

i= (p within r)=2a

p = the voltage of the ui electromotive force.

e=pTotal i=16 2=8v

Terminal voltage u=e-ir=6V

The power of R2 P1 = U R2 = 6W

So the power of the branch with bulb l p2 = p out - p1 = 6w so r lamp + r1 = r2

R-lamp = 4 2) R-lamp: R2=2:1

Therefore, to make the bulb shine normally, the voltage added to both ends of the bulb is 4V, so to make the bulb work normally.

r1+r=r lamp.

r1=3 At this time, the total resistance at the end of the road is r'=r lamp + r1=7 current in the circuit i = e (r + r 1 + r lamp) = 1a The output power of the power supply is pout'=i r'=1 7=7w total power p total'=i r total'=1²×8=8w

Efficiency of the power supply = p out'p total'=7/8×100%=

1. Calculate the current in the main circuit of the circuit first There is a problem to know that the power consumed on the electromotive force is 4w p=i r to get the main current to be 2a

In this way, the external circuit is considered as a whole, and the resistance is r, then the output power is 12=i r and the push r is 3

Therefore, the resistance of RL in series with R1 in parallel with R2 is 3 ohms.

Launched with an RL of 4

2. According to the first question, the power supply electromotive force can be calculated because the total power of the electromotive force is 16w = ui and the launch e = 8v

Or according to the first problem, the voltage at both ends of the sub-circuit is 6V, that is, the RL and R1 are also 6V, and the voltage division is known as UL=4V

That is, the rated voltage of the bulb, which is 4V, is 1A

When the switch S is disconnected, if you want L to shine normally, you must ensure that the circuit current is 1A

Since the power supply electromotive force is 8V, the total resistance must be 8

All the time the internal resistance of the power supply r=1 the internal resistance of the bulb 4 Therefore, the resistance of the rheostat should be 3 ohm bulb to work normally.

In this case, the rheostat resistance is 2 ohms, so p should be slid to the right to r1 = 3 ohms.

The output power of the power supply is EI=8x1=8W, while the consumption on the bulb and rheostat is IR 1x7=7W

So the efficiency is 7 8x100%=

1) The power of the power supply internal resistance r pr=p total p transmission = i 2 * r = 16-12 = 4, substitute r = 1, get i = 2a, the internal resistance is divided to the voltage ur = ir = 2 * 1 = 2v, the voltage at the end of the road is the parallel circuit voltage u and = p total i-ur = 6v, i (r2) = u and r2 = 6 6 = 1a, by the shunt of the parallel circuit, i (r1) = i - i (r2) = 2-1 = 1a, then the first branch resistance r3 = u and i (r1) = 6 ohms, then the lamp resistance r (l) = r3-r1 = 6-2 = 4 ohms. (2) From (1), the normal luminous current of the lamp i(rl)=1a, at this time, r total = u total i(r3) = 8 ohms, then r1 = r-r(l)-r = 8-4-1 = 3 ohms, at this time, the internal resistance partial voltage of the power supply ur = i (l) * r = 1 * 1 = 1v, efficiency = [(u total - ur) u total] * 100% = 7 8 =

The resistance of the bulb is 4.

R1 reaches its maximum and the bulb glows normally.

1.When I was in junior high school physics, I often thought that the voltage of the power supply circuit was constant. Do you still see it that way? Under what circumstances is that view approximately correct?

2.In both circuits, the electromotive force of the power supply is the same, but the internal resistance is different. When the current in the circuit is the same, which circuit has the highest road-end voltage?

3.Connect three resistor wires with different resistances in series to connect them to the power supply. If the point is connected to the positive pole of the power supply, and the point is connected to the negative pole of the power supply, how should the points be arranged in the order of potential reduction?

Which two bucket beams have the largest potential drop between the buried points? Which resistance line has the least potential drop?

4.The electromotive force of the power supply in the circuit is , the internal resistance is , and the resistance of the external circuit is . Find the current in the circuit.

5.Connect a fixed-value resistor to the power supply into a circuit as shown in the figure of this question, and the fixed-value resistance is . When the switch is closed, the voltmeter reads . When turned on, the voltmeter reads . Find the internal resistance of the power supply.

6.In this diagram, when the switches are closed, what is each? When the empty ants are disconnected, how much are they each?

7.Given an ammeter, a voltmeter, and a sliding rheostat, how do you measure the electromotive force and internal resistance of a power supply? Draw a circuit diagram that illustrates the method of measurement and calculation.

8.The electromotive force of a certain battery is , the internal resistance is , and when it is connected with the external circuit, the power released by the external circuit is , and the road-end voltage and power supply efficiency of the battery are found.

9.When a battery pack with an electromotive force of is sending current to the slag key of the automobile motor, its end-of-road voltage is . Just ask

1) What is the internal resistance of the battery pack?

2) How much is the chemical energy of the battery pack reduced? How much electrical energy does it output to an electric motor? How much heat is generated inside the battery pack?

Coulomb force: f=kq1q2 r square.

Electric field strength: e=f q e=u d e=kq r.

1, (1) None (2) None (3) Yes.

2. Yes The magnetic flux has changed.

3. There is an induced current in the process of the coil entering and out, and there is no induced current completely inside and completely outside4. The coil moves away from the wire and the coil has an induced current. When the current in the wire gradually increases or decreases, there is also a change in the induced current magnetic flux.

5. The copper ring is a uniform magnetic field, and the copper ring does not induce current. There is no change in the magnetic flux. There is no uniform magnetic field on the copper ring, and the copper ring has an induced current. The magnetic flux has changed.

6. In the case of B, C and D, B has induced current, and the magnetic flux has changed7. B0S1=BS2 that is, B0*L*L=B*L*(L+VT) can be solved to obtain B=B0*L (L+VT).

It's tragic to have no textbooks, and I can't answer you.