Physics in the second year of junior high school, Ohm s law, calculation problems

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The ammeter uses a range, the voltmeter uses a 15V range, the power supply voltage is 36V, R1 is a fixed resistance, R2 is a sliding rheostat, and the resistance of R2 is connected to the circuit.

, the indication of the ammeter is.

Now the current through R1 is changed by adjusting R2, but it must be ensured that the ammeter does not exceed its range, Q:

1) What is the resistance of R1? (2) What is the minimum resistance of the R2 access circuit? (3) When R2 is the minimum value, what is the reading of the voltmeter?

You didn't tell me whether it was in series or in parallel, how do I calculate?

22.The current when the small bulb is glowing normally:

Resistance: 6V (.)

The solution is r=23Resistance of fixed value resistor: 4V

So when u=5v, i=5v8=

Because it is larger, it is not possible to measure the current in this fixed resistance with an ammeter of range.

i= r=(

r=4 i=u r=5 8= "So it is possible to measure the current in this fixed-value resistance with an ammeter.

1 The current passing through the normal operation is ampere.

At a voltage of 6 volts.

i.e. r=ohms (r is the resistance of the series.

2 r=4 ohms.

When the voltage is 5, the current through the resistance is 5 8 = amp is greater than ampere, so the ammeter with a range of range cannot be used.

Solution: Let the resistance of rx be x ohms.

Analysis: Supply voltage = ohms + rx).

rx = ohms + rx).

Depending on the topic, the equation can be listed:

x=solution: x=10

So rx = 10 ohms.

Set the power supply voltage to u

When s is closed: u=

When s is disconnected: u=

From the above equation we get rx=10

u=ar

Build a system of equations:

When S is closed: U=A is closed rx

When S opens: U=A open (Rx+R0).

A closed rx = a open (rx + r0).

5rx = 10 ohms.

（1）i*r=u

Then (2) when slipping down section B, the voltmeter is 2V, then the voltage of the resistor R1 is 1V, then the current is.

The same is true for the current of R2, which is 2V

Then the resistance of R2 = 2V ohms).

In fact, the second question can be done like this: because the voltage ratio of R2 to R1 is 2:1, then the resistance ratio is also 2:1, and it is known that R1=5, so R=10

1.If L1 and L2 are connected in parallel, it is known that the ratio of the currents passing through L1 and L2 is 1:2, and R1:R2

In parallel, the voltage u is the same. r=u/i

So, r1:r2=i2:i1=2:1

2.If L1 and L2 are connected in parallel, the ratio of resistance between L1 and L2 is known to be 4:3, and the current passing through L2 is 2A.

Same as above, i1:i2=r2:r1=3:4

So, i1=3 4i2=3 4*2=3 2a3It is known that the ratio of resistance of R1 and R2 is 3:2, and the ratio of the current passing through them is 4:3, if the voltage at both ends of R1 is, find the voltage at both ends of R2.

Formula: U=IR

u1 u2=(i1r1) (i2r2)=(4*3) (3*2)=2So, u2=u1 2=

1. The parallel voltages are equal, so i1r1=i2r2, that is, r2 r2=i2 i1=2:1

2. The same 1 has i2=i2r2 r1=2*3 4=3 2=

3. u1 u2=i1r1 i1r2=i1 i2*r1 r2=4 3*3 2=2 so u2=u1 2=

1) Parallel connection means that the voltage at both ends of L1L2 is equal, the ratio of resistance = the ratio of voltage divided by the ratio of current The ratio of voltage value is 1:1, and the ratio of current value is 1:2, so the ratio of resistance value is 2:

12) Parallel connection means that the voltage at both ends of L1L2 is equal, the ratio of current = the ratio of voltage divided by the ratio of resistance to the ratio of voltage value is 1:1, and the ratio of resistance value is 4:3, so the ratio of current value is 3:

If the current of 4L2 is 2A, then the current of L1 is 3 2A

3) The ratio of voltage = the ratio of resistance multiplied by the ratio of current = 3 2 * 4 3 = 2: 1r1 voltage is then r2 voltage is.

I seem to have returned it to the teacher, and the children should use their brains more: 1 Parallel circuit, r1 r2 voltage is equal Voltage current = resistance Set the voltage to 1 then 1 1:1 2=2:1

2. In parallel circuits, the voltage of R1 R2 is equal and the current ratio is the inverse of the resistance L1 and L2 current ratio = 3:4 2 4 * 3 =

3. Resistance multiplied by current = voltage r1, r2 voltage ratio is (3*4) :(2*3) = 2:1

1.In a parallel circuit, the voltage is equal, then the resistance is inversely proportional to the current, so r1:r2=2:1

2.L1 is four thirds times the current of L2, and the answer, of course, is eight-thirds amperes. i.e. 8 3a

3.The formula is not good to type out on the computer, just write the answer:

Let the first indication of the voltmeter be u1, the second time is u2, the first indication of the ammeter is i1, the second time is i2 (r1+r2)·i1=u ri+r2 2)·i2=u from i1=u (

1.Because in parallel, the voltage is the same.

l1:l2=1:2

So ir1:r2=2:1

r2=4:3, then i1:i2=3:4 i1=3 4*2a=:r2=3:2 i1:i2=4:3

So u1:u2=(r1*i1):(r2*i2)=2:1u2=u1 2=

(1) is a parallel circuit i1:i2=r2:r1 1:2=2:1 r1:r2=2:1

2) Yes parallel circuit i1:i2=r2:r1 i1:i2=3:4 i2=2a i1=

3)u=ir u1/u2=（i1r1）/（i2r2）=（4*3）/（3*2）=2 ∴u2=u1/2=