. I=U/R Ohm's Law Formula Some circuit Ohm's Law formulas: I=U/R or I = U/R = Gu (I = U: R) formula shows G = 1/R, the reciprocal g of resistance R is called conductance, and the resistance measured by Ohm's Law in the international system of units is Siemens (S). Among them, I, U and R- are the current intensity, voltage and resistance of the same part of the circuit at the same time. I = q/t current = charge amount/time (all units are in the international system of units), that is to say: current = voltage/resistance or voltage = resistance × current "can only be used to calculate voltage and resistance, which does not mean that resistance has a changing relationship with voltage or current" Scope of application Ohm's law applies to metal conduction and electrolyte conduction. Ohm's law will not apply to gas conduction and semiconductor devices. The formula derived from ohm's law is: series circuit: I total =I 1=I2 (in series circuit, currents everywhere are equal) U total =U 1+U2 (in series circuit, The total voltage is equal to the sum of the voltages at both ends of each part) RTotal = r1+R2+R3 ...+RNU1:U2 = r1:R2 (series connection to form a positive partial pressure) P 1/P2=R 1/R2. The main current is equal to the sum of the currents of each branch) U total =U 1=U2 (in parallel circuit, the power supply voltage is equal to the voltage at both ends of each branch)1/rtotal = 65438 (r1+R2) rtotal = r1r2r3: (r/kloc). R 1 when there are n fixed-value resistors R0 connected in parallel, the total resistance R=R0/n, that is, the total resistance is smaller than any branch, but the more parallel connections, the smaller the total resistance. Ohm's law series voltage divider (voltage) parallel shunt (current) partial circuit For any given closed circuit, according to Ohm's law, the current flowing through any resistor multiplied by the resistance value is the voltage across the resistor. The sum of the voltages on all resistors is the electromotive force of the power supply. Because the current direction of the internal circuit is from the negative electrode to the positive electrode, we can think that the power supply voltage is negative. So we come to the conclusion that the algebraic sum of the voltages shared by all the appliances in the closed circuit is zero. From this, it can be inferred that in any complex circuit, any closed circuit can also draw the following conclusion (that is, Ohm's law of some circuits): after a given direction (clockwise or counterclockwise), the algebraic sum of voltages shared by all electrical appliances is zero. Edit the formula I=E/(R+r)=(Ir+U)/(R+r) I- Current Ampere (A) E- EMF Volt (V) R- Resistance Ohm (ω)R- Internal Resistance Ohm (ω)U- Voltage Volt (V). r P release = EI P output = UI P inside = I？ R P output =I? R =E？ R/(R+r)？ =E？ /(R+2r+r？ When r = R, p output is the largest, and p output =E? /4r (mean inequality) (power efficiency is the highest when the output power of the power supply is the largest) Power efficiency n (efficiency) = poutput/lease = iu/ie = u/e = r/(r+r) As can be seen from the above formula, the greater the external resistance r, the higher the power supply efficiency. When R=r, the power supply efficiency is 50. Special case: when the external circuit is disconnected, R=∞, I=0, Ir=0, U = E. That is, the electromotive force of the power supply is numerically equal to the voltage when the external circuit is disconnected. (2) When the external resistance R decreases, according to I=E/(R+r), the current I increases (e and r are fixed values), the internal voltage Ir increases, and according to U=E-Ir, the terminal voltage U decreases. Special case: when external resistance R=0 (short circuit), I=E/r, internal resistance Ir=E and terminal voltage U=0. (Pay attention to prevent short-circuit accidents in practical use) The differential form of ohm's law takes a cylindrical small-volume element in the electrified conductor, whose length is Δ l and cross-sectional area is Δ s, and the axis of the cylinder is along the direction of current density j, then the current Δ i flowing through Δ s is determined by ohm's law, Δ i = j Δ s =-Δ u/r is determined by resistance r = Δ Δ l/Δ s, It is concluded that ohm's law J δ S =-δ U δ S/(ρ δ L) is based on the relationship between electric field strength and potential, and -δ U/δ L = E, then: J = 1/ρ * E = σ E (e is the electric field strength, σ is (2) the resistance of the total resistance of series resistors is equal to the sum of the resistances of all parts, that is, R series = The total resistance of parallel (1) parallel resistors is less than that of any one voltage divider resistor. (2) The reciprocal of the total resistance of the parallel resistor is equal to the sum of the reciprocal of each part, that is, R parallel/1= r11+R2/1+...+rn/1. Edit the formula of ohm's law in this paragraph (including the derivation formula). The main formula comes from ohm's law: parallel circuit I always =I 1+I2 series circuit I always =I 1=I2 ohm's law experiment U always = U1= U2U65438+U2++U65433. +1: R2R total = r1+R2++rni1:I2 = R2: r1u1:U2 = r1:r2r total = r/kloc-0. +0R3) That is to say: current = voltage ÷ resistance or voltage = resistance× current The current flowing through the resistor in the circuit is directly proportional to the voltage applied across the resistor and inversely proportional to the resistance value of the resistor. (1) Series Circuit P (electric power) U (voltage) I (current) W (electric power) R (resistance) T (time) Current is equal everywhere I 1=I2=I The total voltage is equal to the sum of the voltages at both ends of all appliances U=U 1+U2 The total resistance is equal to the sum of all resistances R. Sum w = w1+w2w1:w2 = r1:R2 = u1:u2p1:P2 = r1:R2 = u1. U=U 1=U2。 The reciprocal of the total resistance is equal to the sum of the reciprocal of each resistance. R = r 1r 2(r 1+R2) Note: This is only limited to two parallel resistors. If there are multiple resistors, the reciprocal of the equivalent resistance of the total circuit is equal to the sum of the reciprocal of each branch resistance, and the total electric work is equal to the sum of each electric work W = w1+W2i1:I2 = R2: r1w1:W2 = I1:I2. The sum of power P=P 1+P2 Ohm's Law (3) The electric power of the same appliance ① The rated power is equal to the square Pe/Ps=(Ue/Us) compared with the actual voltage. Related circuit formula ① Resistance R = ρ L/S Note: ρ is not density, but length 1m, and the cross-sectional area is1mm 2. The resistance of wire material at room temperature ② is equal to voltage divided by current R = u÷êI ③ is equal to voltage square divided by electric power R = u÷êp ê electric work W is equal to current multiplied by voltage. Electric work equals electric power times time W=Pt, and electric work equals charge times voltage W=QU. Square resistance times time W=I×IRt (pure resistance circuit). The electric power is equal to the voltage squared divided by the resistance, and then multiplied by the time w = u u÷r× t (same as above). (3) electric power P 1 equals voltage times current P = UI 2 equals current squared times resistance P=IIR (pure resistance circuit). (3) The electric power is equal to the voltage square divided by the resistance p = P=UU÷R (same as above). (4) Electric power equals electric power divided by time p = w: TT (4) Electric heating Q equals current squared into resistance multiplied by time Q=IIRt (universal formula) Electric heating equals current multiplied by voltage multiplied by time Q=UIt=W (pure resistance circuit) Circuit change of Ohm's law 1. Problems related to circuit change can be divided into (1) the problem of judging the change of electric representation (. (2) Range selection and change division of electric energy meter; (3) Range of sliding rheostat. Second, it can appear in the form of filling in the blanks, selecting and calculating. Third, the analysis method: (1) See clearly the wiring mode of the circuit before and after the change, how the access resistance changes due to the movement of the sliding vane of the sliding rheostat, and connect and disconnect the changed circuit. First, see clearly what wiring mode the circuit is before and after the change; (2) From the circuit diagram, analyze which part of the circuit current and voltage are measured by ammeter and voltmeter respectively; (3) According to the nature and characteristics of the series-parallel circuit, use ohm's law to solve problems flexibly. & lt/FONT & gt； & lt/SPAN>。