Question

1. in a high pass rl filter, the output voltage increases as ________. a resistance increases b frequency increases c frequency decreases d current decreases

Explanation:

In a high - pass RL filter, the output voltage is taken across the resistor. The impedance of the inductor \(Z_L = j\omega L=j2\pi fL\) (where \(f\) is the frequency). As the frequency \(f\) increases, the impedance of the inductor \(Z_L\) increases. Using the voltage divider formula \(V_{out}=\frac{R}{R + Z_L}V_{in}\), when \(Z_L\) increases (due to increasing \(f\)), the denominator \(R + Z_L\) is dominated less by \(R\). But for a high - pass filter, as \(f\) increases, the inductor acts more like an open - circuit at very high frequencies. Mathematically, the transfer function \(H(f)=\frac{R}{\sqrt{R^{2}+(2\pi fL)^{2}}}\). As \(f\) increases, \(H(f)\) approaches \(1\) (since \(2\pi fL\gg R\) at high \(f\)), so \(V_{out}\) approaches \(V_{in}\).

• Option a: Resistance is a fixed element (assuming no variable resistor in the basic analysis), and changing \(R\) (if it were variable) would not be the characteristic behavior of a high - pass filter based on its frequency - dependent operation.
• Option c: If frequency decreases, \(Z_L = 2\pi fL\) decreases. Using the voltage divider \(V_{out}=\frac{R}{R + Z_L}V_{in}\), as \(Z_L\) decreases (denominator \(R+Z_L\) gets smaller, but for a high - pass filter, we want \(V_{out}\) to be large for high frequencies). When \(f\) decreases, \(V_{out}\) decreases (low - pass like behavior).
• Option d: Current \(I=\frac{V_{in}}{R + Z_L}\). As \(V_{out}=IR\), if current decreases (either because \(V_{in}\) changes or \(R + Z_L\) changes in a non - frequency - related way which is not the essence of a high - pass filter's frequency - dependent operation), it doesn't explain the high - pass behavior.

Answer:

B. frequency increases