As shown in the figure, the coil with area s, number of turns n, resistance R and resistance R form a closed loop. The ideal AC voltmeter is connected in parallel at both ends of resistance R. the coil rotates uniformly at an angular velocity ω around the rotation axis perpendicular to the magnetic field in a uniform magnetic field with magnetic induction intensity B. suppose the time when the coil rotates to the position shown in the figure t = 0 A. When t = 0, the coil is in the neutral plane, the current flowing through the resistance R is 0, and the reading of the voltmeter is 0b. Within 1 second, the direction of the current flowing through the resistance R changes ω π times C. after connecting another resistance at both ends of the resistance R, the reading of the voltmeter will decrease D After a capacitor with larger capacitance is paralleled at both ends of the resistance R, the reading of the voltmeter will not change, and the signal processing system will count every time it gets a high voltage

As shown in the figure, the coil with area s, number of turns n, resistance R and resistance R form a closed loop. The ideal AC voltmeter is connected in parallel at both ends of resistance R. the coil rotates uniformly at an angular velocity ω around the rotation axis perpendicular to the magnetic field in a uniform magnetic field with magnetic induction intensity B. suppose the time when the coil rotates to the position shown in the figure t = 0 A. When t = 0, the coil is in the neutral plane, the current flowing through the resistance R is 0, and the reading of the voltmeter is 0b. Within 1 second, the direction of the current flowing through the resistance R changes ω π times C. after connecting another resistance at both ends of the resistance R, the reading of the voltmeter will decrease D After a capacitor with larger capacitance is paralleled at both ends of the resistance R, the reading of the voltmeter will not change, and the signal processing system will count every time it gets a high voltage

A. When t = 0, the coil is in the neutral plane, the magnetic flux through the coil is the largest, the electromotive force is 0, the current is 0, but the reading of the voltmeter is the effective value of the voltage, the effective value is not 0, so the reading of the voltmeter is not 0, so a is wrong; B, the cycle of the alternating current is t = 2 π ω, the direction of the alternating current changes twice in a cycle, so it flows through the resistance R in 1 second The number of times of changing the direction of current is 2 × 1t = ω π, so B is correct; C. after a resistor is paralleled at both ends of resistance R, the total resistance will decrease, the total current will increase, and the internal voltage of coil will increase. As the total electromotive force remains unchanged, the external voltage, that is, the reading of voltmeter will decrease, so C is correct; D. the capacitor has the characteristics of AC and DC isolation, and is paralleled at both ends of resistance R After connecting a capacitor with larger capacitance, the alternating current can pass through the capacitor, so the total resistance decreases. According to the analysis of C, the reading of voltmeter will decrease, so D is wrong

If the magnetic flux passing through a closed coil with an internal resistance of 1 Euclidean decreases by 2 Webers per second, the induced electromotive force in coil a increases by 2 volts per second, the induced current in coil b decreases by 2 amperes per second, there is no induced electromotive force in coil C, and the induced current in coil D remains unchanged, equal to 2 amperes

I'll answer it for you! Please accept it in time! The answer is: D because the topic is about uniform reduction, not acceleration or deceleration, then the electromotive force remains unchanged, then the current remains unchanged!

When the magnetic flux in the closed coil changes uniformly, the direction of the induced electromotive force in the coil will change
A closed coil is located in a uniform magnetic field, and its magnetic field intensity changes at a uniform speed. Suppose you take any two points in the coil and ask about the potential between them.
(the biggest problem is that we can't judge which part of the conductor is the power supply.)

The electric field excited by the changing magnetic field in the surrounding space is called the vortex electric field. The vortex electric field is a non conservative field, and its electric field line is a closed curve with no beginning and no end. The potential energy cannot be defined by the non conservative field

The resistance of a uniform resistance wire is 20 ohm. Fold it in half and twist it together to form a strand. How many ohm is its resistance

According to r = PL / S
It can be seen that the smaller the length and the larger the cross-sectional area, the smaller the resistance
After folding, the cross-sectional area is doubled and the length is half
So l / S is a quarter of the original
It is 5 ohm

When the resistance in a certain circuit increases by 4 Ω, the current will decrease by 1 / 5 of the original value. How many Ω should the resistance be increased
Please, my physics is a mess
The voltage does not change

According to u = IR, the original resistance is R and the original current is I
Using the same voltage, we can get the following results
IR=（R+4）x1/5 I
The solution is: r = 1

After the resistance of a section of conductor is increased by 3 Ω, it is connected to the original power supply, and it is found that the current of the conductor is 4 / 5 of the original current?

Let the original resistance be r and the current be I
IR=4/5I(R+3)
The solution is r = 12 Ω
The original resistance of the conductor is 12 Ω

When the resistance in a circuit is increased by 8 Ω and the current is reduced to 1 / 3 of the original, what is the original resistance

Because I = u / R, I / 3 = u / (R + 8), and,
I / u = 1 / r = 3 / (R + 8)
R + 8 = 3R
R=4
A: the original resistance was 4 ohm

The resistance of two resistance wires is 2 Ω and 16 Ω respectively. If they are connected in parallel at both ends of the power supply, what is the ratio of the current passing through the gate?
It's better to have a picture. You can describe what the picture looks like

I=U/R
I1/I2=R2/R1=8:1

The resistance of two resistance wires is 2 ohm and 16 ohm respectively. If they are connected to both ends of the same power supply, the ratio of current passing through them is I1: I2=___ ．

Suppose the voltage of the power supply is u, then when two resistance wires are connected respectively, according to Ohm's law, the ratio of current passing through is: I1: I2 = ur1: UR2 = R2: R1 = 16 Ω: 2 Ω = 8:1

A resistance wire with uniform thickness is connected to a 6 V power supply. The current passing through it is 0.2 a, and the negative value of the resistance wire is () Ω. If the resistance wire is folded in half and connected to the power supply, the current value of the wire passing through the resistance is () a

30 Ω, 0.4A 6 divided by 0.2 = 30, 0.2 times 2 = 0.4