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Why not a step up transformer be used to increase the power generated many times and solve energy crisis?
Because a transformer does not generate power, it transformers it. Power is equivalent to the voltage times current. A transformer with a ratio of N1:N2 takes voltage (V1) and current (I1) at one winding and transforms it into (N2/N1)*V1 voltage and (N1/N2)*I1 current at the other winding. So the input power is V1 * I1, and the output power is (N2/N1)*V1*(N1/N2)*I1 = V1*I1 (ignoring the small amount of losses associated with the transformer).
what are the advantage of current transformer?
These are used with low range ammeters to measure current in HVAC 's where direct connection of instruments is impratical. They not only insulate the instruments from HV lines but also step down current in a known ratio. ( i .e . ) Iline =(I1 \I2) * Ammeter Reading where Iline = Line Current I1 \I2 = Current Ratio
WAP to interchange the value of two variable without third variable?
Ellipses (...) used to emulate indentation... swap (int *i1, int *i2) { /* only works for integers, i1 != i2 */ ... *i1 = *i1 ^ *i2; ... *i2 = *i1 ^ *i2; ... *i1 = *i1 ^ *i2; }
What is the Amperes difference between a 2 light bulb series circuit and 2 light bulb parallel circuit?
if the bulbs are identical, we can assume that both of them have the same resistance R. in series : the flowing current is I1, the voltage at the ends of the 1st Bulb is V1 and the voltage at the ends of the 2nd bulb is V2. V1 = V2 = V since they have the same resistance and V is the voltage applied on the ends of the whole circuit. the Total circuit resistance is 2*R. thus : I1 = V / 2*R => V = 2 * I1 * R (1) in Parallel mode : the current flowing in the first bulb is i1 and the current flowing in the second bulb is i2. i1 = i2 since R1 = R2. I2 = i1 + i2 i1 = V / R and i2 = V / R => I2 = (V /R ) + (V / R) = 2*V / R => V = I2 * R / 2 (2) (1) and (2) => 2 * I1 * R = I2 * R /2 => I1 = I2 * R / (R * 2) => I1 = I2 / 2 above, is the mathematical solution of the problem, but the result is directly predictable since both bulbs are identical. in summary the current that flows in the circuit when wired in series is half of the current when wired in parallel. also note, that the voltage in series on the ends of each bulb is half of the voltage when wired in parallel.
How do currents flow round two series and parallel circuits?
In a series circuit, the current does not change. It is simple the total voltage divided by the total impedence. However, in a parallel circuit the current splits. More current will go down the path with less impedence, and vice versa. For example, you have two parallel branches, 1 and 2, with resistances R1 and R2 respectively. Also, the total current flowing is I, while the current down each branch is I1 and I2. I1= I(R2)/(R1+R2) I2= I(R1)/(R1+R2) Also note that I1 + I2 = I.
How do you calculate primary curuent?
Current is calculated on the load. If your question on transformer primary current, then use the formula N1I1=N2I2, where N1 and N2 are primary and secondary coil turns and I1 and I2 are current in respective coils. This is very basic simple formula. You have reframe your question more specifically.
How do the current and voltage in the individual resistors compare when resistors are connected in series?
When "n" number of varying resistances are connected in series R total = R1+R2+R3+R4+ . . . . . = Req V total = (V1* I1)+(V2* I1)+(V3* I1)+(V4* I1) { As I1=I2=I3=I4} V total = V battery I total = V battery / Req = I1=I2=I3=I4= . . . . . = Ieq
Why not a step up transformer be used to increase the power generated many times and solve energy crisis?
Because a transformer does not generate power, it transformers it. Power is equivalent to the voltage times current. A transformer with a ratio of N1:N2 takes voltage (V1) and current (I1) at one winding and transforms it into (N2/N1)*V1 voltage and (N1/N2)*I1 current at the other winding. So the input power is V1 * I1, and the output power is (N2/N1)*V1*(N1/N2)*I1 = V1*I1 (ignoring the small amount of losses associated with the transformer).
what are the advantage of current transformer?
These are used with low range ammeters to measure current in HVAC 's where direct connection of instruments is impratical. They not only insulate the instruments from HV lines but also step down current in a known ratio. ( i .e . ) Iline =(I1 \I2) * Ammeter Reading where Iline = Line Current I1 \I2 = Current Ratio
WAP to interchange the value of two variable without third variable?
Ellipses (...) used to emulate indentation... swap (int *i1, int *i2) { /* only works for integers, i1 != i2 */ ... *i1 = *i1 ^ *i2; ... *i2 = *i1 ^ *i2; ... *i1 = *i1 ^ *i2; }
What is the Amperes difference between a 2 light bulb series circuit and 2 light bulb parallel circuit?
if the bulbs are identical, we can assume that both of them have the same resistance R. in series : the flowing current is I1, the voltage at the ends of the 1st Bulb is V1 and the voltage at the ends of the 2nd bulb is V2. V1 = V2 = V since they have the same resistance and V is the voltage applied on the ends of the whole circuit. the Total circuit resistance is 2*R. thus : I1 = V / 2*R => V = 2 * I1 * R (1) in Parallel mode : the current flowing in the first bulb is i1 and the current flowing in the second bulb is i2. i1 = i2 since R1 = R2. I2 = i1 + i2 i1 = V / R and i2 = V / R => I2 = (V /R ) + (V / R) = 2*V / R => V = I2 * R / 2 (2) (1) and (2) => 2 * I1 * R = I2 * R /2 => I1 = I2 * R / (R * 2) => I1 = I2 / 2 above, is the mathematical solution of the problem, but the result is directly predictable since both bulbs are identical. in summary the current that flows in the circuit when wired in series is half of the current when wired in parallel. also note, that the voltage in series on the ends of each bulb is half of the voltage when wired in parallel.
How do currents flow round two series and parallel circuits?
In a series circuit, the current does not change. It is simple the total voltage divided by the total impedence. However, in a parallel circuit the current splits. More current will go down the path with less impedence, and vice versa. For example, you have two parallel branches, 1 and 2, with resistances R1 and R2 respectively. Also, the total current flowing is I, while the current down each branch is I1 and I2. I1= I(R2)/(R1+R2) I2= I(R1)/(R1+R2) Also note that I1 + I2 = I.
What is i to the sixth power?
The powers of i are: i1 = i i2 = -1 i1 = -i i1 = 1 After that, the pattern repeats, so i6 = -1.
What is the current value of the Browning 12 gauge B25 I1 serial 8l3RP7536?
100-1500 USD depending on specifics
How Core Loss is dependent on Frequency?
since at no load only excitation current(responsible for core loss ie iron loss) flow on the primary side so core loss current will be 1A and core loss = v1*i1*powerfactor. core loss = 1*11000*0.24= 2640watt.
Is current ratio of transformer constant on loads?
the current ratio I1/I2 is a constant for all loads. the total current drawn on the secondary winding of a transformer depends on the total circuit impedance as seen by the transformer output terminals. considering that the power in primary is equal to the power in the secondary and is a constant (assuming no losses as per the principle of conservation of energy). The input voltage(primary) is a constant i.e equal to the primary distribution line voltage, and the secondary distribution line voltage is also operated at a constant value. therefore KVA(primary) = KVA( secondary) V1I1=V2I2 I1/I2 =V2/V1 which is a constant irrespective of the load.
How do you calculate turns ratio for current transformer?
The turns ratio is the number of primary turns divided by the number of secondary turns. This is the same ratio as input current to output current. ie the turns ratio N = I1/I2
