What is the relationship between the current ratio, the power ratio and the resistance ratio in a series circuit

What is the relationship between the current ratio, the power ratio and the resistance ratio in a series circuit

In a series circuit, the current is equal everywhere,
So the total current: I = I1 = I2, so I1: I2 = 1:1 has nothing to do with the resistance
Voltage U1 = i1r1, U2 = i2r2, so U1: U2 = i1r1: i2r2 = R1: R2
Power P1 = I1 ^ 2r1, P2 = I2 ^ 2r2, so P1: P2 = I1 ^ 2r1: I2 ^ 2r2 = R1: R2
The area of the triangle is 3cm square. Try to write the functional relationship between the height y (CM) and the bottom x (CM)
xy/2=3,
y=6/x,
Is an inverse proportional function
Y=6\X
Triangle area s = bottom * height / 2
∴S=XY/2=3
So y = 6 / X (x > 0)
The sum of the first n terms of sequence {an} is Sn, A1 = 1, an + 1 = 2Sn + 1 (n ≥ 1), and the general term formula of {an} is obtained
From an + 1 = 2Sn + 1 (n ≥ 1), we can get an = 2sn-1 + 1. By subtracting the two formulas, we can get an + 1-an = 2 (sn-sn-1) = 2An (n ≥ 2), an + 1 = 3an (n ≥ 2), and a 2 = 2S1 + 1 = 3 = 3A1, so the sequence & nbsp; {an} is equal ratio sequence, and the common ratio is 3, the first term is 1
How to prove the relationship between electric power and resistance
The existing power supply voltage, an ammeter, a voltmeter, a variety of known resistance, a number of wires, a number of verification: when the current is constant, the resistance power is proportional to the resistance
Upstairs, the size of the two parallel resistance is not right, it should be channeled connected
Two resistors, ammeters and batteries of different sizes are connected by leaps and bounds, and the current flowing through the two resistors is the same. Measure the voltage at both ends of the two resistors respectively (the voltage with larger resistance is higher). According to P = IIR, calculate P1 = iir1, P2 = IIR2
Then P1: P2 = iir1: IIR2, that is, P1: P2 = R1: R2 is provable~
The ammeter is connected in parallel with the large and small resistors, and the voltage on the two resistors is tested respectively. It can be seen that the voltage shared by the large resistor is high, and the voltage shared by the small resistor is low,
Yes,
1. The current of the two measurements does not change, and the resistance does not change. From u = I * r, we can see that the voltage on both sides of the resistance does not change.
2。 From P = u * I. The measured value can be brought into the formula R1 / r2 = P1 / P2, and it is found that the equation holds. It can be seen that the resistance power is proportional to the resistance when the current is constant. ... unfold
The ammeter is connected in parallel with the large and small resistors, and the voltage on the two resistors is tested respectively. It can be seen that the voltage shared by the large resistor is high, and the voltage shared by the small resistor is low,
Yes,
1. The current of the two measurements does not change, and the resistance does not change. From u = I * r, we can see that the voltage on both sides of the resistance does not change.
2。 From P = u * I. The measured value can be brought into the formula R1 / r2 = P1 / P2, and it is found that the equation holds. It can be seen that the resistance power is proportional to the resistance when the current is constant. Put it away
When the area of a triangle is 10cm ∧ 2, the functional relation between its bottom edge a (CM) and its height h (CM) is----
When the area of a triangle is 10cm ∧ 2, the functional relation between its bottom edge a (CM) and its height h (CM) is a = 20 / h
The sum of the first n terms of the sequence {an} is Sn, A1 = 1, an + 1 = 2Sn + 1 (n ≥ 1) (1) find the general term formula of {an}; (2) the items of the arithmetic sequence {BN} are positive, the sum of the first n terms is TN, and T3 = 15, and a1 + B1, A2 + B2, A3 + B3 are proportional sequence, find TN
(1) Because an + 1 = 2Sn + 1 ① So an = 2sn-1 + 1 (n ≥ 2) ② So (1) by subtracting the two formulas, we can get an + 1-an = 2An, that is, an + 1 = 3an (n ≥ 2). Because A2 = 2S1 + 1 = 3, so A2 = 3A1, so {an} is an equal ratio sequence with the first term of 1 and the common ratio of 3. Let {BN} have a tolerance of D. from T3 = 15, we can get B1 + B2 + B3 = 15 and B2 = 5, so B1 = 5-D, B3 = 5 + D, and because A1 = 1, A2 = 3, A3 = 9, and a1 + B1, A2 + B2, A3 + B3 are equal ratio sequence We can get (5-D + 1) (5 + D + 9) = (5 + 3) 2, and get D1 = 2, D2 = - 10 ∵ the items of the arithmetic sequence {BN} are positive, ∵ d > 0, ∵ d = 2, ∵ TN = 3N + n (n − 1) 2 × 2 = N2 + 2n
On the calculation of electric power and thermal power in high school
First, P = UI is introduced, all resistors can be used, and then thermal power is calculated according to u = IR, q = w = I R. I am very curious. Since I R is introduced according to u = IR, then u r can also be introduced. Why can't u r be used to calculate thermal power?
Is it because the general topic is given to 220 V, rather than the voltage on the required resistance? However, since the topic says that the rated voltage is 220 V, the required resistance should be 220
U-r can be used to calculate the thermal power if it is a pure resistance circuit, because the formula u = IR is a pure resistance circuit
P = UI is not only suitable for resistance, but also for electrical appliances
Ohm's law applies only to resistance
Q = I2R is also applicable to all electrical appliances
In the circuit composed of pure resistance, the current has only thermal effect, w = q and Ohm's law u = IR is applicable,
Therefore, in the circuit composed of pure resistance, w = q = IU = I ^ 2 * r = u ^ 2 / R
(where u is the voltage at both ends of R and I is the current passing through R)
Use of pure resistance circuit
The functional relationship between the length a (CM) of the bottom edge of an isosceles triangle with an area of 60cm ^ 2 and the height h (CM) of the edge is?
h=60*2／a
Given that the sum of the first n terms of sequence an is Sn, A1 = 2, Nan + 1 = Sn + n (n + 1), (1) find the general term formula of sequence an; (2) let BN = sn2n, if BN ≤ t exists for all positive integers n, find the minimum value of T
(1) ∵ Nan + 1 = Sn + n (n + 1) ∵ (n-1) an = sn-1 + n (n-1) (n ≥ 2) can be obtained by subtracting Nan + 1 - (n-1) an = sn-sn-1 + 2n, that is, Nan + 1 - (n-1) an = an + 2n, (n ≥ 2) can be obtained by sorting, an + 1 = an + 2 (n ≥ 2) (*) can be obtained by A1 = 2, A2 = S1 + 2 = 4, a2-a1 = 2 is suitable for (*). Therefore, the sequence {an} is an arithmetic sequence with 2 as the first term and 2 as the tolerance, and can be obtained by the general term formula of arithmetic sequence , an = 2 + (n-1) × 2 = 2n (2) can be obtained from (1), Sn = n (n + 1), ｜ BN = sn2n = n (n + 1) 2n from the monotonicity of the sequence, BK ≥ BK + 1, BK ≥ bk-1k (K + 1) 2K ≥ (K + 2) (K + 1) 2K + 1K (K + 1) 2K ≥ K (K − 1) 2K − 1. Solving the inequality, we can get 2 ≤ K ≤ 3, K ∈ n *, k = 2, or K = 3, B2 = B3 = 32 is the maximum term of the sequence {BN}, and t ≥ 32 can be obtained from BN ≤ t
The ratio of the electric power of the series parallel circuit to the heat generated in the same time
Such as the title
Power is always equal to UI
In a storage resistor circuit, the power is equal to the square of I multiplied by R or the square ratio of u r
Heating is to calculate the power consumed by the resistance in the circuit. The formula is the same as above, I square times R or u square ratio R
Series parallel circuit you want to grasp is in the series circuit current is equal, at this time should use the formula I square multiplied by R, parallel point out to grasp the voltage is equal, to use the formula u square ratio R
In fact, the two formulas are equivalent!
Ratio of electric power of series parallel circuit: R1 * R2
-------
(R1+R2)
The ratio of heat produced is equal to the ratio of electric power
Constant current of series circuit (power P = I * I * r)
Parallel circuit voltage unchanged (P = u * U / R)
But the resistance is not the same. The resistance in series is equal to the sum of the resistances in the trunk line, and the resistance in parallel is not the same
We have to calculate the resistance.