If different bulbs are connected to the same circuit, whether the current in this circuit changes

Conjecture: If different bulbs are connected to the same circuit, the current will be **.

Equipment: wires, ammeters, bulbs of different powers.

Experimental procedure: Probably according to different connections: series and parallel, and then use an ammeter to measure the current at both ends of each bulb.

It will change, the internal resistance of different bulbs is different, so the resistance in the loop will also change, when the total voltage is unchanged, the loop resistance changes, the current changes, the power changes, the experimental equipment: switch 1 ammeter 1 voltage source 1 bulb several, guide several.

Experimental steps: 1 Connect the wire according to the circuit diagram, 2 Close the switch, record the current indicator, and observe the change of the brightness of the bulb.

3. Record the current I1 and various data, 4. Replace the circuit diagram and repeat the above steps.

5. Analyze the data and conclude that the current will change with the change of the circuit connected to different bulbs.

6. Fill in the experiment report.

7. Pack up the equipment and put it away according to the regulations.

Of course, there are changes, the current changes according to the different loads, and if you want to do the experiment, you need multiple light bulbs, the same number of switches and ammeters, 3 things in series, connected to the battery in parallel, close the different switches, and you can see the change of current.

One, in series, two, in parallel.

Third, series and parallel mixing.

Equipment: small beads (small bulbs), wires, switches, battery packs, ammeters, voltmeters.

Experiment with series circuits.

Procedure, experiment with parallel circuits.

Procedure, experiment with series and parallel circuits. Step 1,

i=p uIf the voltage of the same circuit does not change, the power of different bulbs changes, and the current also changes.

To put it simply, 2 bulbs must have a larger current than 1 bulb, and when using 1 bulb, the current is smaller, and when you connect another bulb to increase the current, it must be connected in parallel (series connection will increase the resistance in the circuit). Test apparatus, omitted.

If there is a change in the resistance value, there will be a change.

<> solution: 1. When USB1 singularity clears and separates, USB2 and US are short-circuited, above.

3 The voltage at both ends is: U=US1 (3 Acacia before 2) (3 2+2)=18.

So: i'=u/3=。

2. When US acts alone, US1 and US2 are short-circuited, as shown in the figure above.

i"=-us/（3+2∥2）=-7/（3+1）=-7/4（a）。

3. When us2 acts alone, us1 and us are short-circuited, as shown in the figure above.

u=us2××（2∥3）/（2∥3+2）=12×。

So: i'"=u/3=。

4. Superposition theorem: i=i'+i"+i'"=。

Summary. Hello, in a simple battery and bulb circuit, the magnitude of the current entering the bulb from the side close to the positive side of the battery and leaving the bulb from the opposite side is equal. This is because according to Kirchhoff's law of current, the sum of currents at any node in a circuit is equal to zero, i.e., the currents are continuous in the circuit.

Therefore, the current enters the bulb from the positive end of the battery, creating a certain energy draw and voltage drop in the bulb, and then returns to the negative end of the battery from the other end of the bulb so that the current in the circuit can be balanced.

In a simple battery and bulb circuit, is the current entering the bulb from the side close to the positive side of the battery greater than leaving the bulb from the opposite side.

Hello, in a simple battery and bulb circuit, the magnitude of the current entering the bulb from the side close to the positive side of the battery and the current leaving the bulb from the opposite side are equal. This is because according to Kirchhoff's current law, the sum of currents at any node in the circuit is equal to zero, i.e., the currents are continuous in the circuit. Therefore, the current enters the bulb from the positive end of the battery, creating a certain amount of energy consumption and voltage drop in the bulb, and then returns from the other end of the bulb to the negative end of the battery, so that the current in the circuit is known to be balanced.

Can you ask any other questions?

OK. If we reduce the potential difference between resistors in a circuit, the current flowing through that resistor increases, stays, or decreases.

According to Ohm's law, the relationship between the resistance r in a circuit and the deficit current i is i = v r, where v is the voltage across the resistor. If we reduce the potential difference between the resistors in the circuit, i.e., the voltage v across the resistors, then according to Ohm's law, the current i will decrease. This is because in a guess circuit, the current is driven by the potential difference, and the greater the potential difference, the greater the current, and vice versa.

Therefore, if the potential difference decreases, the current also decreases.

The current will decrease.

1. Three bulbs are connected in series, and the voltage at both ends of each lamp is U lamp = 12V 3 = 4v When the image knows that U lamp = 4V, I lamp =, so R lamp = = 4V ohm.

2) Let the current through the bulb be i, and the voltage at both ends of the bulb is u, then there is u total = 2ir0 + u, and u = 8-20i to make the image of u = 8-20i can be seen that the curve in the intersection diagram is in (2, point, as shown in the figure current representation number i table = then the resistance is 20 3 ohms.

I've been reading it for a long time, and I still can't understand your problem.

Can you make the topic clearer?

That might help you out.

Let the DC voltage be U

Option A, S1 is closed, S2 is off, capacitor C is equivalent to an open circuit, and the current flowing through L1 is U (R1+R2).

When S2 is closed, the current flowing through L1 is U (R1+R2 R3), so the current increases and L1 becomes brighter.

Option B, after S2 is closed, L2 is connected in parallel with L3, so the resistance decreases, and the distributed voltage also decreases, so UL2 decreases and L2 is dimmed.

When option C, S1 is closed, and S2 is off, the voltage of the left plate of capacitor C is U, and the voltage of the right plate is 0

When S1 is closed and S2 is closed, the voltage of the left plate is U, and the voltage of the right plate is greater than 0, so the power decreases.

Option D, due to i=c*(du dt), when the voltage change rate is 0, du dt=0. Now the voltage across the capacitor changes from U to UL1, so du dt <0, so the current direction is from right to left.

The answer is c

Solution, connect a bulb to both ends of the power supply with constant voltage, and the electric power of the bulb is 40 watts, that is, 40 = U square R lamp··· 1）

After connecting the bulb and a certain resistor r in series, the electrical power of the resistor is watts, that is, the square * r = (u (r lamp + r)) square * r · 2)

Solution (1) (2) formula R lamp = 1 9r or R lamp = 9R (not in line with the topic, discard) At this time, the power of the bulb p = i square * r lamp = (u (r lamp + r)) square * r lamp, r lamp = 1 9r substitute and combine it.

1) The formula gives p=

Since the bulb resistance is less than r, the bulb power must be greater than it, and it is unlikely to be reached, so answer A should be chosen.

Let the voltage be U and the bulb resistance is R, according to Ohm's law, the total current of one bulb i=U R, and connect two small bulbs in parallel, then the resistance is r 2, in the same way, according to Ohm's law, the total current of two bulbs i=2u r

The latter is twice the current of the former.

Yes, the total resistance is reduced by parallel connection, and the current is increased if the voltage does not change.

If the bulbs are the same, the current of the latter is greater than that of the former.

If the small bulbs are different, the situation is complicated.

If the bulbs are the same, it's right.

The latter has twice as much current. The point current of each branch is the same as that of the former.

Yes, the former has only one small bulb's current, while the latter is the sum of the currents of two small bulbs (in the case of the same bulb).