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Chapter 32: Q. 42 (page 925)

For an RC filter circuit, show that $V_{R} = V_{C} = \frac{蔚_{0}}{\sqrt{2}}$ at $蝇 = 蝇_{c}$ .

Short Answer

Expert verified

The expression is true $V_{R} = V_{C} = \frac{蔚_{0}}{\sqrt{2}}$ .

Step by step solution

01

Given Information

We are given that $蝇 = 蝇_{c}$ and

we have to find out that $V_{R} = V_{C} = \frac{蔚_{o}}{\sqrt{2}}$ .

02

Simplification

As we know that for an RC filter circuit,

$蝇_{c} = \frac{1}{R C}$

Next we will find V_R :

$V_{R} = \frac{蔚_{0} R}{\sqrt{R^{2} + \frac{R^{2} C^{2}}{C^{2}}}}$

We get , $V_{R} = \frac{蔚_{0}}{\sqrt{2}}$

Finally, we will find V_C:

$V_{c} = \frac{\frac{蔚_{0}}{蝇_{c}}}{\sqrt{R^{2} + \frac{1}{蝇^{2} C^{2}}}}$

$V_{C} = \frac{蔚_{0}}{\sqrt{蝇^{2} R^{2} C^{2} + 1}}$

We get, $V_{C} = \frac{蔚_{0}}{\sqrt{2}}$

As $V_{R} = V_{C} = \frac{蔚_{o}}{\sqrt{2}}$

Hence, it is proved that $V_{R} = V_{C} = \frac{蔚_{0}}{\sqrt{2}}$ .

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Most popular questions from this chapter

a. For an RC circuit, find an expression for the angular frequency at which $V_{C} = \frac{1}{2} 蔚_{0}$ .

b. What is V_R at this frequency ?

A high-pass filter consists of a $1.59 m F$ capacitor in series with a $100 鈩�$ resistor. The circuit is driven by an $A C$ source with a peak voltage of $5.00 V$ .

a. What is the crossover frequency $f_{c}$ ?

b. What is $V_{R}$ when $f = \frac{1}{2} f_{c}, f_{c}$ , and $2 f_{c}$ ?

For the circuit of figure $E x 32.33$ ,

a. What is the resonance frequency, in both $r a d / s$ and $H z$ ?

b. Find $V_{r}$ and $V_{c}$ at resonance.

c. How can $V_{c}$ be larger than $蔚_{0}$ ? Explain .

a. Evaluate $V_{R}$ in figure at emf frequencies $100, 300, 1000, 3000, 10000 H z$

b. Graph $V_{R}$ versus frequency. Draw a smooth curve through your five points.

A $110 H z$ source of emf has a peak voltage of $50 V$ . Draw the emf phasor at $t = 3.0 m s$ .

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