The reading for both the DC ammeter and the AC ammeter in a half-wave rectifier circuit is given by 2sinwt A. The total input power in this circuit is 11000 sin^2(wt) W.
In a half-wave rectifier circuit, the DC ammeter is connected in series with the load resistance, while the AC ammeter is connected in parallel with the rectifier resistance. In this case, we have a rectifier resistance of 500Ω and a load resistance of 4500Ω.
To find the reading for each ammeter, we need to calculate the current flowing through both resistances. The current flowing through the rectifier resistance can be found using Ohm's Law: I = V/R, where V is the voltage and R is the resistance. In this case, the voltage is given as V = 1000sinwt. Substituting the given values, we get I = (1000sinwt) / 500 = 2sinwt A.
The reading for the DC ammeter, which is in series with the load resistance, is the same as the current flowing through the load resistance. Thus, the reading for the DC ammeter is 2sinwt A. The reading for the AC ammeter, which is in parallel with the rectifier resistance, is the same as the current flowing through the rectifier resistance. Thus, the reading for the AC ammeter is also 2sinwt A.
To calculate the total input power, we need to find the power dissipated by both the rectifier resistance and the load resistance. The power dissipated by a resistor can be found using the equation: P = I^2 * R. Substituting the given values, we get the power dissipated by the rectifier resistance as P = (2sinwt)^2 * 500 = 2000 sin^2(wt) W. The power dissipated by the load resistance can be calculated as P = (2sinwt)^2 * 4500 = 9000 sin^2(wt) W. The total input power is the sum of the power dissipated by both resistances, so the total input power is P_total = 2000 sin^2(wt) + 9000 sin^2(wt) = 11000 sin^2(wt) W.