High School Physics, Electrical, Closed Circuit Analysis

B, C, when P slides to the resistance of the two branches is equal, the resistance of the parallel circuit is maximum, and U reaches the maximum value, that is, the voltage at both ends of the parallel circuit increases first and then decreases, C pairs.

In the process of increasing to the maximum value, the resistance of the R1 branch decreases, the number of current representations increases, greater than I, the voltage increases to the maximum value and then slides upwards P, the total resistance of the circuit decreases, the current in the dry circuit increases, with the decrease of the resistance value of the parallel circuit, the voltage of the parallel circuit decreases, and the resistance at the bottom of the P point increases and the current decreases, so the current in R1 should increase, so the current in R1 is always greater than the I and B pairs.

When P slides to the two branches with equal resistance, the resistance of the parallel circuit is maximum, and U reaches the maximum value A is wrong. The voltage at both ends of the parallel circuit increases and decreases first, in the process of increasing to the maximum value, the resistance of the R1 branch decreases, the number of current representations increases, that is, greater than I, the voltage increases to the maximum value and then slides upwards P, the total resistance decreases, the current in the trunk begins to increase, the voltage of the parallel circuit decreases, and the current in the resistance below the P point decreases, so the current in R1 needs to increase urgently, so the current in R1 is always greater than I, B pair. At the same time, it can be seen that C is true and D is false.

Select BC

The voltage divider circuit has internal resistance and starts with the current ......

When the motor does not rotate, the coil only heats up, Ohm's law applies, and the resistance of the motor coil.

r=u/i=

The electrical power of the motor when it is working normally.

p=u'i'=9v×

The power lost by the motor coil in heat generation.

P loss = i'²r=(

The power output of the electric motor.

P machine = p - p loss =

Electric motor efficiency.

p machine p=

1, E is the ideal power supply, that is, E has no internal resistance, A1 and A2 are connected in parallel, the voltage at both ends of A1 and A2 is equal to E, then A1 remains unchanged, A2 becomes larger, and the total current becomes larger;

2, E is not an ideal power supply, that is, E has internal resistance, A1 and A2 are connected in parallel, A1 and A2 are regarded as a whole and then connected in series with the internal resistance of the power supply, R2 becomes smaller, A1 and A2 overall internal resistance becomes smaller, the total resistance of the power supply does not change, the total current becomes larger, the voltage at both ends of the power supply internal resistance becomes larger, the voltage at both ends of A1 and A2 becomes smaller, A1 becomes smaller, and the total current becomes equal to A1 + A2, and the total current becomes larger, so A2 becomes larger, that is, A1 becomes smaller, A2 becomes larger, and the total current becomes larger.

Look at the conditions given to you in the question.

The shunt voltage of the parallel circuit is the total voltage, because the voltage does not change, the resistance value of R1 does not change, so A1 does not change.

When R2 becomes smaller, i=U, r A2 becomes larger, i=i1+i2 so the total current becomes larger.

Firstly, without considering the internal resistance of the power supply, the voltages of the R2 and R1 branches remain unchanged, so R2 decreases and the A2 current increases. The R1 current does not change. Therefore, the total current is the sum of the two currents, and the total current increases.

Since R1 is a fixed resistance, the current of A1 does not change when R2 decreases. When R2 becomes smaller, the current becomes larger, and A2 becomes larger, because the total current is equal to the current of each branch, so the current of A1 remains the same, and A2 becomes larger, so the total current becomes larger.

A1 does not change, A2 becomes larger, and the total current also increases.

In parallel circuits, the voltages at both ends of the two resistors are equal.

According to the title, it can be obtained that r1 * a1 = r2 * a2 = e the total voltage e does not change, when the switch s is closed, r2 becomes smaller, a2 becomes larger, a1 does not change, the total current = a1 + a2, so the total current also becomes larger. Please adopt.

By p=i 2r=e 2r (r+r) 2=e 2r [(r-r) 2+4rr] when r=r, the output work reed collapse rate is up to 1 pair.

The gear of the power supply is only efficient =iu ie=r (r+r)=1 (1+r r) r large high efficiency 2 wrong.

When sliding upward, the resistance of the sliding rheostat increases and the RP increases.

Then the resistance of R2 in series with P and then in parallel with R1 increases, then the voltage obtained by this part increases, so the voltage of L1 increases, variable.

Since P has to be divided with L2, and its effect is far more awesome than the effect of partial pressure just now, then I won't write the specific formula for L2 darkening, and the computer is too troublesome.

B because of the series inverse merger: the resistance in series with the resistor that changes the resistance value in addition to its resistance value does not change, its voltage, current and electrical power will be opposite to the change direction of the changed resistance (the resistance value of the sliding rheostat becomes smaller, L2 and its series connection will reduce the voltage, current, and electrical power of L2, and the brightness of L2 will be effective), and the parallel connection will be the same as its change direction.

This can be used for most topics.

In the process of moving the dicing from A to B, the resistance of the access circuit increases first and then decreases, then the total current of the circuit decreases first and then increases, and the total power of the power supply P=EI decreases first and then increases, C pairs; If the current decreases, the internal resistance of the power supply decreases, and the voltmeter reading (external circuit voltage division) increases first and then decreases, and D is wrong; The charge of the capacitor Q=UC, and when the current decreases, the voltage at both ends of RN decreases, resulting in Q decreasing, so the charge is first decreasing and then increasing, B pair.

Correct answer BC

Select B, C, Solution: In the process of sliding rheostat from A to B, the resistance value of the access circuit increases first and then decreases, and the resistance value of the sliding rheostat is the largest when it slides to the middle. The voltmeter measures the voltage of the external circuit, so it should be increased first and decreased, and D is wrong.

The capacitor and the stationary resistor are connected in parallel, so it is determined by the voltage on both sides of the stationary resistor, so it should be given whether it is reduced first and then increased, and A is wrong and B is right. The internal resistance and external resistance of the file are the same as the maximum power, the greater the difference, the smaller the power, so the greater the resistance value of the external circuit, the smaller the power, and the resistance value of the external circuit is increased first and then decreased, so it is first reduced and then increased.

It's been years since I graduated from high school, and I hope my solution is right......

1, The output power of the power supply p out=4W

2. The electrical power lost inside the power supply p=2W

3. The efficiency of the power supply =4 6=

The resistance of the voltmeter v2 is R2 and the internal resistance is R

e=22+130+130r r2 1 formula.

e=150+150r r2 2 formula r r2=1 10

Substituting 2 or 1, e=165v

Big brother, no one will give you the money for questions without bounty points.