A source with an impedance of 100 $\Omega$ is connected to the primary coil of a transformer and a resistance R is connected to the secondary coil.If the transformer has 500 turns in its primary coil and 100 turns in its secondary coil the greatest power will be dissipated in the resistor if R =
A)0 $\Omega$
B)0.25 $\Omega$
C)4.0 $\Omega$
D)20 $\Omega$
E)100$\Omega$

Question

Quizplus · Accepted Answer

The impedance seen by the secondary coil of the transformer is given by Z_sec = Z_prim*(N_sec/N_prim)^2, where Z_prim is the impedance of the source, N_sec is the number of turns in the secondary coil, and N_prim is the number of turns in the primary coil. Substituting the given values, we get Z_sec = 100*(100/500)^2 = 1 F1F1F1S1F1F1F10.
The power dissipated in the resistor is given by P = V^2/R, where V is the voltage across the resistor. The voltage across the resistor is given by V = I*R, where I is the current in the circuit. The current in the circuit is given by I = V/Z_sec. Substituting the values, we get I = 10 F1F1F1V1F1F1F1S1F1F1F11 = 10 F1F1F1A1. Therefore, the power dissipated in the resistor is P = (10 F1F1F1V1F1F1F1S1F1F1F11)^2/R = 1000/R. For maximum power dissipation in the resistor, we need to minimize R. Therefore, the answer is the smallest value of R given, which is 4.0 F1F1F1S1F1F1F10.

A Source with an Impedance of 100 Ω\OmegaΩ Is Connected to the Primary Coil of a Transformer and to the Primary

A Source with an Impedance of 100 $\Omega$ Is Connected to the Primary Coil of a Transformer and to the Primary