When the resistance of the conductor remains unchanged, the current flowing through the conductor will increase from 1.5A if the voltage at both ends of the conductor is increased from 4V to 6V A. To 3.5A B. to 2-25a C. to 0.75a D. to 7.5A

When the resistance of the conductor remains unchanged, the current flowing through the conductor will increase from 1.5A if the voltage at both ends of the conductor is increased from 4V to 6V A. To 3.5A B. to 2-25a C. to 0.75a D. to 7.5A

When the voltage is 4V, the resistance R = 4 / 1.5 = 2.67 Ω is calculated. When the voltage is 6V, the current I = V / r = 6 / 2.67 = 2.25A, so B is selected

Calculation: (√ 6 + √ 5) & # 178; - (√ 6 - √ 5) & # 178;

Original formula = [(√ 6 + √ 5) + (√ 6 - √ 5)] [(√ 6 + √ 5) - (√ 6 - √ 5)]
=(2√6)(2√5)
=4√30

In the experiment of measuring the electric power of a small bulb, what happens after the ammeter and voltmeter are switched off?

Ammeter string in the circuit, no doubt
1. The voltmeter is connected to both ends of the power supply. When the ammeter and the voltmeter exchange positions, the power supply is short circuited. One of the power supply or the ammeter is burned first, and the voltmeter has no indication
2. The voltmeter is connected to both ends of the bulb. When the ammeter and the voltmeter exchange positions, the bulb is short circuited, the ammeter hardly shows the number, and the voltage is the power supply voltage

How to solve equation 1.8x = x + 6 * 2

Move the right x to the left
1.8x-x=12
Digong factor
(1.8-1)x=12
0.8x=12
x=15

As shown in the figure, the total voltage of the circuit remains unchanged. When the switch S3 is disconnected and only the switch S1 and only the switch S2 are closed respectively, the power ratio of resistance R1 and R2 is 4:3, and the voltage ratio of both ends of resistance R3 is 2:3; when only the switch S2 is closed, the electric power consumed by resistance R3 is 9W

Is this it?

Calculation: (1 + x) (1-x) (1 + x ^ 2) (1 + x ^ 4)

Original = (1-x & # 178;) (1 + X & # 178;) (1 + x ^ 4)
=(1-x^4)(1+x^4)
=1-x^8

Electrical power is 1 kilowatt. How many kilowatts of electricity can I use for an hour

Theoretically, it is one degree power, one degree power is equal to 1kW * 1H, but in fact, many factors need to be considered. For example, in the household power, the voltage may be unstable and low-voltage. At this time, if it is not an electrical appliance under constant voltage, it cannot be calculated with rated power
Don't think too much when doing the questions. If you encounter such questions, the test point is the physical meaning of one degree of electricity. It is the power consumption of one kilowatt electrical appliance working for one hour

It is calculated that 150-120 △ 16 × 0.84 is equal to?

Multiply and divide first, then add and subtract
Original formula = 150-7.5 × 0.84
=150-6.3
=143.7

How to calculate the capacity of lithium battery

The unit of capacity of a lithium battery is MAH, not MAH
If your battery is 1000 MAH, which means that the working current is 1 MAH, the battery can work for 1000 hours. If the working current is 1000 MAH, it can only work for 1 hour. The larger the "MAH", the greater the capacity of the battery

5 out of 13 times 4 out of 7 times 14 is easy to calculate!

5/13x4/7x14
=5/13x(4/7x14)
=5/13x8
=40/13