When ten bulbs are connected in series and in parallel, what is the total resistance ratio of the tw

If the influence of bulb current and temperature on the resistance is not taken into account (i.e., the bulb is considered as a purely resistive load), one bulb resistor is r, 10 in series is 10r, and 10 in parallel is r 10, so the resistance value ratio of the two is 100:1

When the ten bulbs are connected in series and parallel, the total resistance ratio of the two is 100:1

The voltage of each branch in the parallel circuit is the same, so we want to compare the power of the bulb in the parallel circuit by the formula p=u 2 r. The lower the resistance, the more powerful (and brighter) the bulb

In the series circuit, the currents are equal everywhere, so the following formula is used to compare the power of the series circuit: p=i 2*rSo the greater the resistance, the greater the power (the brighter).

1. First of all, the total voltage is set to u

2. Then the two bulbs connected in parallel are regarded as a whole, and the bulb resistance is R1 and R2 respectively, then the total resistance R3 of the two bulbs is R1R2 R1+R2 (according to the parallel resistance formula).

3. The resistance of the set value resistance is R4, and the total resistance of the whole circuit is R3+R44, that is, the total current i is U R3+R4 (according to I=U R).

Total current = current on the resistor, or total current = sum of the currents of the two bulbs.

According to the formula "1 r1 + 1 r2 = 1 r total" for calculating the resistance of the parallel circuit, the total resistance of the two bulbs is calculated by adding the resistance next to it, and the total current is calculated with "i=u r".

Calculate the parallel resistance first plus the series resistance.

S closure is, the resistance is short-circuited, the bulb is glowing brightly, that is, U=6V, the bulb resistance R=6*6 3=12 ohms.

When S is disconnected, the resistor and the bulb are connected in series, because the terminal voltage is unchanged, the electric bi gear voltage at both ends of the bulb = 6*12 (12+6)=4V

So the actual power of the bulb = 4*4 12=4 3=

When the S is closed and hail, the resistor is shortened, and the bulb is connected to both ends of the power supply separately.

At this time, the bulb is normal, then U = U = 6V

The resistance of the bulb.

R light = u amount p amount = 6 * 6 3 = 12 euros.

When the s are disconnected, the resistor and the bulb are connected in series.

r total = r resistance + r lamp = 6 + 12 = 18 ohms.

The current of the circuit at this time.

i = u r total = 6v 18 ohm = 1 3a

The actual power of the bulb at this time.

P light = i r light = (Friction 1 3a) * 12 ohms =

--- does not need to calculate, the total power is [200W], because no matter how you connect the parallel resistor on [Bulb 1], there is only one purpose, that is, the total resistance of [Bulb 1] and the parallel resistor is equal to the internal resistance of [Bulb 2]. Only if their resistance is equal, the divided voltage will be equal, and the divided voltage will be equal, and the power consumed will be equal, so the total power is equal to 200W.

--The internal resistance of Bulb 1 = 110V squared divided by 60W = -Bulb 2] Internal Resistance = 110V squared divided by 100W = 121

--Required shunt resistance and power = 40W or more.

From ·p u r to r u p

From ·p u r to r u p

Since the voltages are equal when connected in parallel, the voltage of each resistor does not change, so p and p do not change.

So: p always p p

When connected in series, due to the relationship between the voltage division, the power of the resistor also changes, and it is no longer p and p, so the total power is directly calculated.

p total u r r u u p u p p

The current i flowing in series is the square of (P1 R1) or the square of (P2 R2);

When connected in parallel, the voltage U is (R1+R2) multiplied by the current I, and the power of each bulb can be calculated after knowing the voltage, and the sum of power consumption is the total power.

The voltage of the voltage of the series, current i, r1. r2xi=。ixu2＝p2。。u1^2/r1=。u1+u2=u

In parallel, R1XR2 R1+R2=Total U2R