Physics in the second year of elementary school, the application of Ohm s law, solving, and finding

1. The current range is, the voltage is 9V, R1=6 ohms.

According to Ohm's law, the greater the r, the smaller the current, so the maximum value of r2 is 100 ohms.

Refer to this solution.

According to Ohm's law and the characteristics of series circuits, r1+r2=u i=9, so the minimum value of r2 is 15-6=9

2. First, the use of series voltage division: the voltage of the voltmeter UV=UR1 (R1+R2) when U is the smallest, R2 is 100 maximum; When u is maximum, r2 is the smallest, at this time.

R1+R2=UR1 UV=9*6 3=18 Second, the use of series voltage division: the voltage of the voltmeter UV=UR2 (R1+R2) when U is the smallest, R2 is the minimum 0; When u is maximum, r2 is maximum, at this time.

r1+r2=ur2 uv substitution can be solved r2=r1 2=3

1.When the current in the circuit obtains the maximum value, R2 obtains the minimum value, when i=, r total = 9 ohms = 15 ohms, so R2 minimum = 15-6 = 9 ohms, so 9 ohms

A household has a total of 10 lamps connected in parallel on the circuit, in order to minimize the total current in the circuit then should be (3) 1The lights are all on 2Wait for all to turn off 3Turn on only the lights 4Turn on two crack resistant lamps.

Resistors R1 and R2 are connected in parallel on the power supply, and through the ratio of their current i1:i2=3:1, then their resistance ratio is only 1:3 If they are connected in series on a unified power supply, the ratio of the voltages at both ends of them: 1:3

The resistance of the two wires is r, and after being connected in series, they are connected to both ends of the power supply with voltage U1, and after parallel Zen connection, they are connected to the other power supply at both ends of the power supply with voltage U2. If the series current is equal to the parallel current, then u1:u2 is:4:1

The resistance r1 is 18 ohms, and the maximum allowable current is, and the resistance r2 is 12 ohms, and the maximum allowable current is. After connecting them in parallel, they are allowed to be connected at a maximum voltage of (6)?In the circuit of v, the currents in r1 and r2 are () a

The resistance values of R1 and R2 are 4 ohs and 12 ohms respectively, and they are connected in parallel in a 3V circuit, and the total resistance after parallel connection: (3 ohms) The current through the dry circuit is (1a) R1 The ratio of the current in R2 (3:1).

Switch closing situation: there is no current in the R3 circuit, and U1 measures the voltage at both ends of R2.

Columnable equation: u1

Switch disconnection situation: U1'R2 plus R3 voltage is measured, and U2 is R3 voltage.

Columnable equation: (r2+r3)*'

r3*20+r2+r3)*

Plus the condition u1:u1'：u2=4：9：6

It can be solved that u=24v

r2=20ω

(1) When the switch is placed in A, R1 and R2 are connected in parallel, and R3 is broken. The voltage at R1 and R2 is U=6V, and the current of R1 is U R1=1A.

2) When the switch is placed in B, only R1 is connected in the circuit, R3 is short-circuited, and R2 is open-circuited. The current of R1 is U R1=1A

The voltage on R1 is always 6V, so the current is 1A.

Solution: (1) When S1 is closed and S2 is disconnected, the resistors R1 and R2 are short-circuited, and only R3 works.

The slider is at the far right end again, and the resistance of 10 is connected, and the total resistance r=12; That is, the current is i=u r=6v 12 = is the real number of the ammeter.

The slider is connected to the resistor of 10 at the far right, so the voltage is expressed as u = i * r =.

2) S1 is disconnected, S2 is closed, R2, R3 are short-circuited, only R1 works.

The slide of the rheostat is on the far left, so the sliding rheostat is not connected to the circuit.

The total resistance is r = 9; That is, the current is expressed as i = u r =6v 9 =

The voltmeter is short-circuited by a sliding rheostat, so the indication is 0V.

Thank you

I'll give you a breakdown.

Because the voltage is the same everywhere in the parallel circuit, and the sum of the currents of each branch is equal to the total current, the power supply voltage can be known by only finding the R2 voltage.

When closing s, the current of R1 and L2 is measured, so , so L1=, all the time L2 resistance is 2, and the current is , so 2 is equal to, so L2 voltage, so total voltage.

Find the resistance value of R2, the voltage divided by L1 is equal to R1, the voltage of R1 is always , the current is , so it is equal to 6 So, the total voltage (supply voltage) is , and the resistance value of R1 is 6

Simply set the power supply voltage to U R1, the resistance value of R1 and the resistance value of R2 to R2U R1+U R2=

u r2= again r2=2 u= u r1= brings u= into r1=6 ohms.

1.If the voltage at both ends of a conductor is 9 volts and the current passing through it is 600 mA, then its resistance is (r=u i=9v). If the voltage applied to it becomes 6 volts, then its resistance is (unchanged, resistance is a property of the conductor itself, it is still 15).

If the current passing through it is amperes, then its resistance is (unchanged, still 15), the voltage at its ends is (u=ir=15), if the voltage applied to it is zero, its resistance is (unchanged, still 15), and the current passing through it is (i=u r=0v 15=0a).

2.Solution: The voltage at both ends of R1 u1=u-u-lamp=4v-3v=1vi1=u1 r1=1v2=lamp.

R lamp = U lamp I lamp = 3V

So the voltage across R1 is (1) volts, and the resistance of the bulb is (6) ohms.

Ohms, 15 ohms, 15 ohms, 9v, 15 ohms, 0

v, 6 ohms.

15 ohms, 15 ohms, 15 ohms, 9v, 15 ohms, 0 resistance is an intrinsic property of conductors.

1v, 6 ohms It's easy to calculate with Ohm's law.

In the first question, the option is not stated in the case of a certain voltage; b option: When the voltage is constant, the current in the conductor is inversely proportional to the resistance of the conductor; The c option does not state that it is the same resistor.

Question 2, 90 euros. From the question, we can see that u=u2-u1=9v, i=i2-i1=. According to Ohm's law, r = u i = 90 ohms.

Question 3, b. The current through the two resistors in a series circuit is equal, and if the resistance of r A is 2r and the resistance of r B is r, then i = u (r A + r B) = u 3r. Then U A = i*r A = 2 3*U, U B = i * r B = 1 3 * U, so U A = 2u B.

1. D ohms.

Analysis: If the voltage is 3 V, the current through the resistor is i, then there is 3 r=i, 12 r=i+, and r=90 ohms.

3. B analysis: the current everywhere in the series current is equal, so the ratio of the voltage at both ends of the conductor is equal to the ratio of the resistance of the conductor.

A is not right, such as r is big, but u is also big, i is not necessarily small.

b is not true, when the voltage u is the same, the current through the conductor with a large r is small.

c Pair. d right.

io=uo/r=3/r

i=u/r=12/r

12/r -3/r=9/r=i-io=

r=9 ohms.

Series equicurrent i

UI:u2=IR1:IR2=R1:R2=2:1

It's a little forgotten, but it can still help you push it out.

From Ohm's law r*i = 4*i*x;

As the current increases, the total resistance decreases. Therefore, the method of parallel connection is adopted.

r*x /(r+x)) = r/4;

can be pushed to r 4 = 3*x 4;

The final result is x = r 3

where * denotes multiplication and denotes division.

If the voltage remains constant, then the current is i=u r

If it is a parallel resistor, then there is 1 r (resistance after parallel) = 1 r + 1 (1 3r) = 4 r, so r (and) = 4 r, and then substitute Ohm's law to obtain the current is 4 times the original, that is, 4i

Because i=u r

The power supply voltage remains the same, and his current becomes 4i, indicating that the current becomes larger, and his number will not change until the previous resistor is connected in parallel.

ir=4ir'The title says that the voltage remains the same.

r'=1/4r

1/4r=（r*r'')/(r+r'')

So calculate r''=1/3r。