1. The voltage at both ends of the circuit is 6 V, the resistance of R1 is 8 Ω, and the current through L1 is 0.2A Solution: 1. Current through L2. 2. Voltage at both ends of L1. 3. Voltage at both ends of L2. 4. Resistance of L2 2. R1 and R2 are connected in series in a circuit with a voltage of 6 v. the current passing through the resistor R1 is 0.2 a, R2 = 10 Ω, and the resistance of R1 is calculated The first question is series connection

1. The voltage at both ends of the circuit is 6 V, the resistance of R1 is 8 Ω, and the current through L1 is 0.2A Solution: 1. Current through L2. 2. Voltage at both ends of L1. 3. Voltage at both ends of L2. 4. Resistance of L2 2. R1 and R2 are connected in series in a circuit with a voltage of 6 v. the current passing through the resistor R1 is 0.2 a, R2 = 10 Ω, and the resistance of R1 is calculated The first question is series connection

I2 = I1 = 0.2A the current through L2 is 0.2au1 = I1 × R1 = 8 × 0.2 = 1.6V, the voltage at both ends of L1 is 1.6V, U2 = utotal-u1 = 6-1.6 = 4.4V, the voltage at both ends of L2 is 4.4V, R2 = U2 / I2 = 4.4 / 0.2 = 22Ω, the resistance at L2 is 22Ω, the resistance at R1 = u / i-r2 = 6 / 0.2-10 = 30-10 = 20Ω, the resistance at R1 is 20Ω

In a certain circuit, the voltage at both ends of the conductor remains unchanged. If the resistance of the conductor is increased by 8 Ω, the current will become three fourths of the original current, and the resistance in the original circuit will be three fourths______ Ω．

Suppose the power supply voltage is u, the original resistance is r, and the original current is 34i. According to the application of Ohm's law, we can get the following solution: u = IRU = 34i (R + 8 Ω): r = 24 Ω, so the answer is: 24 Ω

A student used an experiment to study the relationship between the resistance of a conductor and the length, cross-sectional area and material of the conductor. The specific method is to connect two different alloy wires a and B to the circuit with the same voltage, and observe the current intensity IA and IB passing through them. The following four phenomena can directly explain that the resistance of a conductor is related to the length of the conductor
A. When a and B are nickel chromium alloy wires of the same thickness and manganese copper wires of the same thickness and a is longer than B, IA < IBB. When a and B are nickel chromium alloy wires of the same length and a is thicker than B, IA > IBC. When a and B are nickel chromium alloy wires of the same thickness and a is longer than B, IA < IBD. When a and B are nickel chromium alloy wires and manganese copper wires of the same thickness and length, IA < IB

A. When a and B are nickel chromium alloy wire and manganese copper wire with the same thickness, and a is longer than B, that is, the material selected is different and the length is different, so the relationship between the resistance and the length of conductor can not be compared, so it is not in line with the meaning of the question; B, when a and B are nickel chromium alloy wire with the same length, and a is thicker than B, IA > IB; that is

P is a point on the right branch of the hyperbola X & # 178 / 9-y & # 178 / 16 = 1, and m n are circles (x 5) &# 178; Y & # 178; = 4 and (x 5) &# 178; Y & # 178; = 4 respectively-
P is the point on the right branch of hyperbola X & # 178 / 9-y & # 178 / 16 = 1, m n is the point on circle (x 5) &# 178; Y & # 178; = 4 and (X-5) &# 178; Y & # 178; = 1 respectively, then the maximum value of pm-pn is obtained

P is a point on the right branch of hyperbola x ^ 2 / 9 - y ^ 2 / 16 = 1, m and N are points on circle (X-5) ^ 2 + y ^ 2 = 4 and circle (x + 5) ^ 2 + y ^ 2 = 1 respectively, then the maximum value of PM | - | PN | is (detailed explanation)

First, the centers of the two circles are the two focuses of the hyperbola
So | po1 | - | PO2 | = 2A = 6
The radius of the left circle is 2
So | PM | max = | po1 | + 2,
The radius of the right circle is 1, so | PM | min = | PO2 | - 1, so | PM | - | PN | = 6 + 2 + 1 = 9

As shown in Figure 1, the straight line y = - 12x + 1 intersects the x-axis at point a and the y-axis at point B. C (m, - M) is a point on the straight line AB, and the hyperbola y = KX passes through point C. (1) find the coordinates of point C and the analytic formula of hyperbola. (2) as shown in Figure 2, take CB as the edge, make a square BCDE above the straight line AB, find the coordinates of point D, and judge whether point D is on the hyperbola in (1)? (3) As shown in Figure 3, m and F are the points on the edges CD and BC of the square BCDE respectively, and MF ‖ BD, take a point K on the extended line of ED, so that DK = De, KM and EF intersect at the point h. It is proved that: ∠ EDH = 2 ∠ bef

(1) Substituting x = m, y = - m into y = - 12x + 1, we get: - M = - 12m + 1, the solution is: M = - 2, then the coordinate of C is (- 2,2), substituting y = KX to get: k = - 4, then the analytic expression of hyperbola is: y = - 4x; (2) in y = - 12x + 1, let x = 0, the solution is: y = 1, then the coordinate of B is (0,1)

It is known that the line AB intersects two coordinate axes at two points a and B, and OA = ob = 1. The point P (a, b) is the first point on y = 1 / 2x

The point P in the first image is perpendicular to PM and X axis, perpendicular to m and PN and Y axis and N, and the two perpendicular lines and straight line AB intersect E and F. when P moves on the hyperbola y = 1 / 2x, the triangle OEF will change accordingly, then whether there is an internal angle that always remains unchanged in the three internal angles of the triangle, if so, please explain the reason and what the angle is
Let P (a, 1 / (2a)), a > 0
AB:x+y-1=0,
PM:x=a,E(a,1-a),
PN:y=1/(2a),F(1-1/(2a),1/(2a)),
The slope of OE K1 = (1-A) / A,
The slope of of K2 = 1 / (2a-1),
tanEOF=|(k2-k1)/(1+k1k2)|
=1,
The angle EOF = 45 ° is calculated

Given that the line y = 2x + B intersects the coordinate axis at two points a and B respectively, if OA + ob + 3, what is B?
Thank you very much

OA+OB=3?
Let x = 0 get y = B, let y = 0 get x = - B / 2
From the meaning of the title, we get | B | + | - B / 2 | = 3 and | - B | = 2

It is known that the left and right focus of the hyperbola x square - y square = 2 is F1, F2. The moving line passing through F2 intersects with the hyperbola and points a and B
(1) If the moving point m satisfies FM vector = F1A vector + F1B vector + f1o vector (o is the coordinate origin), the trajectory equation of M is obtained
(2) Is there a point C on the x-axis so that the CA vector * CB vector is a constant? If so, find out C

① ∵ hyperbolic equation: X & # 178 / 2-y & # 178 / 2 = 1
Easy to get
F1(-2,0) F2（2,0）
Let m (x, y). A (x1, Y1) B (X2, Y2)
I when the slope of straight line passing through F2 does not exist
The straight line is x = 2
A(2,√3) B(2,-√3)
FM vector = F1A vector + F1B vector + f1o vector
That is, (x + 2, y) = (X &; + 2, Y &;) + (X &; + 2, Y &;) + (2,0)
We get x = x &; + X &; + 4 = 8
y=y₁+y₂=0
M(8,0)
I when k exists, let y = K (X-2) be the moving line passing through F 2
Simultaneous y = K (X-2)
x²-y²=1
We obtain (1-k & # 178;) x & # 178; + 4K & # 178; x-4k & # 178; - 2 = 0 (K ≠ ± 1)
x₁+x₂= -4k²/(1-k²) x₁x₂=(-4k²-2)/(1-k²)
And ∵ FM vector = F1A vector + F1B vector + f1o vector
That is, (x + 2, y) = (X &; + 2, Y &;) + (X &; + 2, Y &;) + (2,0)
We obtain x = x & # 8321; + X & # 8322; + 4 = - 2K & # 178; / (1-k & # 178;) + 4 = 8 - 4 / (1-k & # 178;)
y=y₁+y₂=k(x₁-2)+k(x₂-2)=k(x₁+x₂)-4k=-4k³/(1-k²) -4k=(-4k)/(1-k²)
That is 1 / (1-k & # 178;) = Y / (- 4K)
That is, x = 8-4 / (1-k & # 178;) = 8 + Y / K, k = Y / (X-8)
∴x=8- 4/(1-k²)
That is, X-8 = - 4 / (1-k & # 178;) = - 4 / (1-y & # 178; / (X-8) & # 178;) = - 4 (X-8) & # 178; / ((X-8) & # 178; - Y & # 178;)
That is, 1 = - 4 (X-8) / ((X-8) &# 178; - Y & # 178;)
Namely (X-8) &# 178; - Y & # 178; = - 4 (X-8)
It is reduced to X & # 178; - 12x-y & # 178; + 32 = 0
In conclusion, the trajectory equation of m point is X & # 178; - 12x-y & # 178; + 32 = 0
② Suppose there is C
Let C (m, 0)
CA vector * CB vector = P
Then p = (X &; - M) (X &; - M) + Y &; Y &;
=x₁x₂-m(x₁+x₂)+m²+k²(x₁x₂-2(x₁+x₂)+4)
=(-4k²-2)/(1-k²)①+4mk²/(1-k²)⑤+m²+(-4k⁴-2k²)/(1-k²)②+8k⁴/(1-k²) ③+ (4k²-4k⁴)/(1-k²)④
①+②+③+④=(-4k²-2-4k⁴-2k²+8k⁴+4k²-4k⁴)/(1-k²)
=(-2-2k²)/(1-k²)=2- 4/(1-k²)
5 = - 4m + 4m / (1-k & # 178;)
∴P=m²-4m+2 +(4m-4)/(1-k²)
No matter what the value of K is (K ≠ ± 1), P is a constant
∴4m-4=0
The solution is m = 1
C (1,0)

Given the fixed point F1 (- 2,0) F2 (2,0) in the plane satisfying the following conditions, the trajectory of the moving point P is hyperbolic is (a)
A./PF1/-/PF2/=+_3
B./PF1/-/PF2/=+_4
C./PF1/-/PF2/=+_5
D./PF1/^2-/PF2/^2=+_4
Note: Pf1 / medium / is absolute value+_ 3 is positive and negative
I did it by exclusion,

|PF1-PF2|=2a
Because a