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A circuit composed of a resistor, a power supply and an ammeter in series, The indication of the ammeter is 0.12A; if a 10 ohm resistor is connected in series in this circuit, the indication of the ammeter is 0.1A. Calculate the voltage of the power supply and the resistance in the original circuit
Let the original resistance be R1,
U/R1=0.12
U/(R1+10)=0.1
Solving the equations, R1 = 50 Ω, u = 6 v
The internal resistance of ammeter is RA, the electromotive force of power supply is V, and the sum of internal resistance of power supply and conductor is RV.
Determinant!
V=0.12A*（RA+RV）
V = 0.10a * (RA + 10 ohm + RV)
V is an unknown, (RA + RV) is an unknown, two equations, two unknowns, over
Lie equation
V / 0.12 + 10 = V / 0.1
V=6V
Original resistance = 6V / 0.12A = 5 ohm
U/R=0.12
U/(R+10)=0.1
R = 50 Ω, u = 6 V
The original v = IR = 0.1A * 10 = 1V
The original r = V / I = 1V / 0.12A = 8.33 Ω
U = 0.12r1 from U / R1 = 0.12
Substituting u = 0.12r1 into U / (R1 + 10) = 0.1, 012r1 / (R1 + 10) = 0.1, that is, 0.12r1 = 0.1r1 + 1, 0.02r1 = 1, R1 = 50 Ω
U = 0.12r1 = 50 * 0.12 = 6V
Design experiments, a power supply, a switch, a voltmeter (or ammeter) to compare the size of the resistance to be measured. (1) compare with a voltmeter (2) compare with an ammeter
1. Compare with a voltmeter: the power switch and the resistor are connected in series. Use a voltmeter to measure the voltage at both ends of the two resistors. If the voltage value is large, the resistance will be large
U = IR, series I is the same
2. Compare with the ammeter: the power switch and the resistor are connected in parallel. Measure the current of the two resistors with the ammeter respectively. If the current value is small, the resistance will be large, because
R = u / I, parallel u is the same
The resistance can be measured by volt ampere
(1) Connect the power supply, the switch and the resistance to be measured in series, and measure the voltage of the two resistances respectively with a voltmeter. If the voltage is large, the resistance will be large. Because u = IR, I will not change in the series circuit, and the larger R is, the larger u will be
(2) Connect the power supply, switch and resistance to be measured in parallel. Measure the current of two resistances respectively with an ammeter. The resistance with small current is larger, because u = IR. In the parallel circuit, u remains unchanged. The larger R is, the smaller I is
(1) Connect the power supply, the switch and the resistance to be measured in series, and measure the voltage of the two resistances respectively with a voltmeter. If the voltage is large, the resistance will be large. Because u = IR, I will not change in the series circuit, and the larger R is, the larger u will be
(2) Connect the power supply, switch and resistance to be measured in parallel. Measure the current of two resistances with an ammeter. The resistance with small current will be larger, because u = IR. In the parallel circuit, u remains unchanged. The larger R is, the smaller I is
A power supply with electromotive force of 12 V is connected in series with a voltmeter and an ammeter to form a closed loop. If a resistor is connected in parallel with the voltmeter, the reading of the voltmeter will be reduced to 13 and the reading of the ammeter will be increased to 3 times of the original. Calculate the original reading of the voltmeter
Let the original reading of voltmeter and ammeter be u and I respectively, and the internal resistance of power supply and ammeter be R1 and R2 respectively. According to Ohm's law, it is obtained that ε = u + I (R1 + R2) & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ① ε = 13u + 3I (R1 + R2) & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ② Solve the two equations and substitute the value of ε to get u = 9V, so the reading of voltmeter is 9V
As shown in the figure, the power supply voltage remains unchanged, the ammeter (1), the ammeter (2) and the ammeter (3) are ammeters or voltmeters, and the slide P of the sliding rheostat (R2) is at the middle point. When the switch (s) is closed, the three ammeters all show numbers. When the slide P moves to the right, the following statement is correct ()
A. If the resistance is in parallel, then the indication of Table 1 will not change. B. if the resistance is in parallel, then the indication of Table 2 will increase. C. if the resistance is in series, then the indication of Table 1 will decrease. D. if the resistance is in series, then the indication of table 3 will not change
(1) When the resistors are in parallel, ① is the ammeter to measure the total current, ③ is the ammeter to measure the current passing through the sliding rheostat, ② is the voltmeter to measure the voltage at both ends of the sliding rheostat, that is, the power supply voltage; if the power supply voltage does not change, the indication of ② will not change, so B is wrong; if the power supply voltage does not change, when the slide moves to the right, the resistance of the sliding rheostat access circuit will increase and the current will flow When the resistor moves to the right, the total voltage of the transformer will not change, The resistance value of the sliding rheostat resistance access circuit becomes larger, the partial voltage of the sliding rheostat becomes larger, and the indication of ① becomes larger, so C is wrong, so D is selected
Conversion of Ohm
How many kinds of Ohm? How to convert?
1GΩ=1000000000Ω
1MΩ=1000000Ω
1kΩ=1000Ω
What is the physical meaning of 5 joules
Reference 1 joule: the same object passes 1m in the direction of force under the action of 1n force
Joule is suitable for many physical quantities. For example, work, heat, current and so on. In work, it represents the work done with a force of 1n over a horizontal distance. Heat is the unit of internal energy, and current is the electric energy produced by 5W electrified conductor in normal operation for 1s.
How to calculate resistance with known power, voltage and current?
In the circuit composed of resistance, capacitance and inductance, only resistance consumes power
P=UI=I^2*R
R=P/I^2
If the voltage at both ends of the resistor is known, then:
P=UI=U^2/R
R=U^2/P.
18. [10 Wuhu] the rated voltage and power of household appliances are often not reached when working
(1) (2 points) the resistance of the electric kettle is obtained by P = U2 / r r r = U2 / r r r = 220 V 2 / 1210 w = 40 Ω (2) (4 points) t = 3 min = 180 st the electric energy consumed by the electric kettle is w = 180 / 3600 = 0.5 (kW · h) = 1.8 × 105 j the actual electric power of the electric kettle is p = w / T = 1.8 × 105 / 180 = 1000 (W) (3) (2 points)
What is the 11th power of ten?
What is the 11th power ohm of ten? Is it 100000 megaohm?
10 ^ 11 Ω = 10 ^ 5 megohm
In general, it can be regarded as "open circuit"
The eleventh power of ten is a number and the ohm is a unit of resistance. It doesn't matter.
What is Joule heat in physics, and what is the formula for it
Joule is the unit of measurement of energy, which is the unit of measurement of heat in thermodynamics; it is the unit of measurement of the amount of work done by force in mechanics; in electricity, because the current does work on resistance R, it produces heat, which is called Joule heat
In heat: q = C * Δ t * m
In Mechanics: w = FS
In electricity: u = I ^ 2rt
The unit of measurement of energy is the unit of measurement of heat in thermodynamics and the unit of measurement of work done by force in mechanics.
In heat: q = C * Δ t * m
In Mechanics: w = f * * s
In electricity: q = (square of I) * r * t q = ((square of U) / R) * t
The amount of heat absorbed per unit mass of material at unit temperature (/ k).
RELATED INFORMATIONS
- 1. In the circuit shown in the figure, the power supply voltage remains unchanged, R1 = 10 Ω. When the switch S is closed and the slide P of the sliding rheostat is in the middle point, the indication of the ammeter is 0.3A; when the slide P is moved to the rightmost end, the indication of the ammeter is 0.2A. Then the power supply voltage is 0.3A______ 5. The maximum resistance of sliding rheostat is______ Ω．
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