A transformer consists of 275 primary windings and 834 secondary windings. If the potential difference across the primary coil is 25 V, (a) What is the voltage across the secondary coil? (b) what is the effective load resistance of the secondary coil if it is connected across a 125 ohm resistor.
Ok I got part (a) as being 75.82 V...but I am totally confused on part b I don't know what it is asking for or any equation that would work for it! Help please!
I suppose maybe they mean what is the resistance you will see on the primary side.
Now the power on the secondary side is
P = V^2/R = (75.8)^2 / 125
P = 46 watts
Now I suppose we assume a perfect transformer so we put in that 46 watts on the primary side
46 = 25^2/R
R = 625/46 = 13.6 ohms
In part (b), we are asked to find the effective load resistance of the secondary coil, assuming it is connected across a 125-ohm resistor. To solve this, we can use the turns ratio of the transformer.
The turns ratio (N) of a transformer is defined as the ratio of the number of turns in the secondary coil (Ns) to the number of turns in the primary coil (Np). In this case, Ns = 834 and Np = 275.
The turns ratio can be used to relate the voltages across the primary and secondary coils. It states that the ratio of the voltage across the primary coil (Vp) to the voltage across the secondary coil (Vs) is equal to the turns ratio:
N = Ns/Np = Vs/Vp
To find the voltage across the secondary coil (Vs), we can rearrange the equation:
Vs = N * Vp
Using the given values, we have:
Vs = (834/275) * 25 V
Vs = 75.82 V (which matches your answer in part a)
Now, to find the effective load resistance (RL) of the secondary coil, we can use Ohm's law:
RL = Vs^2 / P
Where P is the power dissipated in the load. In this case, the power dissipated is equal to the square of the voltage across the secondary coil divided by the resistance of the load resistor:
RL = Vs^2 / RL
Substituting the known values, we have:
RL = (75.82 V)^2 / 125 Ω
RL = 45.62 Ω
Therefore, the effective load resistance of the secondary coil is approximately 45.62 ohms.
To answer part (b), we need to calculate the effective load resistance of the secondary coil when it is connected to a 125-ohm resistor.
In a transformer, the ratio of voltages is equal to the ratio of the number of windings of the primary and secondary coils. Therefore, we can use the formula:
(Vp/Vs) = (Np/Ns)
where Vp is the potential difference across the primary coil, Vs is the potential difference across the secondary coil, Np is the number of primary windings, and Ns is the number of secondary windings.
In part (a), you correctly found the voltage across the secondary coil as 75.82 V.
Now, to calculate the effective load resistance, we need to use Ohm's Law, which states that the current (I) through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R), so I = V/R.
In this case, the current flowing through the secondary coil will be the same as the current flowing through the 125-ohm resistor.
Since we know the voltage across the secondary coil (75.82 V), and the resistance of the 125-ohm load resistor, we can substitute these values into Ohm's Law to solve for the current:
I = V/R = 75.82 V / 125 Ω = 0.60656 A (approximately)
Now that we have the current, we can use Ohm's Law again to find the effective load resistance:
R_load = V/I = 75.82 V / 0.60656 A = 124.97 Ω (approximately)
Therefore, the effective load resistance of the secondary coil, when connected to a 125-ohm resistor, is approximately 124.97 ohms.