The rated voltage of the small bulb is 2.5V and the resistance is about 10 ohm The current is 0.04 a

The rated voltage of the small bulb is 2.5V and the resistance is about 10 ohm The current is 0.04 a

Rated power P = rated voltage U * rated current I = rated voltage U * (rated voltage U / resistance R) = u ^ 2 / r = 2.5 * 2.5 / 10 = 0.625w

The electromotive force of the power supply is 30V and the internal resistance is 1 Ω. It is connected in series with a bulb with rated voltage of 6V and rated power of 12W and a DC motor with coil resistance of 2 Ω to form a closed circuit, and the bulb lights normally. What is the output power of the power supply?

The small bulb just lights normally, so the current is: I = plul = 126 = 2A; according to Ohm's law of closed circuit, the voltage at both ends of the motor is: u = e-ir-ul = 30-2 × 1-6 = 22V; so the output power of the motor is: P output = ui-i2r = 22 × 2-22 × 2 = 36W; answer: the output mechanical power of the motor is 36W

Using Ohm's law to deduce the relationship between the total resistance R and R1, R2 (in parallel circuit)

According to Ohm's law, if the total voltage is u and the total current is I, then the total resistance R = u / I
For two resistors in parallel, the voltage is equal, both are u, and the current in the resistor is u
I 1＝U / R1 ,I 2＝U / R2
So the total current I = I 1 + I 2 = (U / R1) + (U / r2)
1 / R＝I / U＝[ （U / R1）＋（U / R2）] / U＝（1 / R1）＋（1 / R2）
Note: it is easy to calculate because u is the denominator

How much is the current of the series circuit when a 5 Ω resistor R1 is connected in series with a 15 Ω resistor R2 and connected to a 6V power supply?

∵ resistor R1 is in series with R2; ∵ r = R1 + R2 = 5 Ω + 15 Ω = 20 Ω; according to Ohm's law, I = ur = 6v20 Ω = 0.3A A: the current of this series circuit is 0.3A

It is known that in the circle O, AB and AC are two equal chords perpendicular to each other, OD ⊥ AB, OE ⊥ ac.d and E are perpendicular feet·

Proof: adoe is a rectangle (from the known three angles as right angles to the known quadrangle Adoes as rectangles) as long as we prove that ad = AE, from the known od vertical AB, OE vertical AC, D, e is the perpendicular foot, so e, D are the midpoint of AC, AB respectively, that is AE = AC / 2, ad = AE / 2, and ab = AC, so adoe is a square

It is known that: as shown in the figure, at point O, the strings AB and AC are perpendicular and equal to each other, OD ⊥ AB is D, OE ⊥ AC is e?
Please be more detailed! There is a problem-solving process!

It is proved that: ∵ OD ⊥ AB, OE ⊥ AC, ab = AC ⊥ AE = ad, ∠ AEO = ∠ ADO = 90 degree, and ab ⊥ AC ⊥ ead = 90 degree ⊥ quadrilateral adoe is a square AB, AC is two mutually perpendicular strings, and OD ⊥ AB is in D, OE ⊥ AC is in E, so quadrilateral adoe is a rectangle, and ab = AC, OD ⊥ AB, OE ⊥ AC, so AE = ad (vertical diameter Theorem

It is known that: as shown in the figure, in the circle O, the strings AB and AC are perpendicular and equal to each other, OD ⊥ AB is D, OE ⊥ AC is e

All three angles are 90 degrees, so it's a rectangle
Because od ⊥ AB is D and OE ⊥ AC is e, ad and AE are half of AB and AC respectively. Because AB equals AC, ad = AE is a square (a rectangle with equal adjacent sides is a square)
I'll write the specific steps by myself. I'll just point them out

As shown in Figure 1, in the circle O, if the chord AB is perpendicular to AC, and ab = AC = 10cm, OD is perpendicular to AB and OE is perpendicular to AC and E, what is the radius of the circle O?

Because the two chords AB and AC are perpendicular, and a is on the circumference
So ∠ BAC = 90,
So the arc corresponding to ∠ BAC is 180
So the BC line goes through the origin, which is the diameter of the circle
So r = D / 2 = (√ (AB ^ 2 + AC ^ 2)) / 2 = (√ (100 + 100)) / 2 = (√ 200 / 2) = 5 √ 2

If od = 3, OE = 4, then the chord AB = () chord AC = () radius OA = ()

Chord AB = (8) chord AC = (6) radius OA = (5)

As shown in the figure, in ⊙ o, the chord ab ⊥ AC, and ab = AC = 2cm, OD ⊥ AB, OE ⊥ AC, the perpendicular feet are D and e respectively, then the inferior arc length of AB is ()
A. 3π2cmB. 3π4cmC. 2π3cmD. 2π2cm

As shown in the figure: connect Ao, Bo ∵ ab ⊥ AC, OE ⊥ AC, OD ⊥ AB, ∵ adoe is rectangular. ∵ AB = AC = 2, ∵ ad = AE = 1, ∵ adoe is square. ∵ Ao = 2, ∠ AOB = 90 °, ab = 90 π 2180 = 2 π & nbsp; 2cm