In the circuit shown in figure the maximum output voltage V0 is

Question

In the circuit shown in figure the maximum output voltage V0 is (A) 0 V (B) 5 V (C) 10 V (D) (5/√2) V

Tardigrade Admin · Accepted Answer

For the positive half cycle of input the resulting network is shown below <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/b10c7c8ab9b09d823938cba88a849193-.png" /> ⇒ (V0) textmax =(1/2)(Vi) textmax = (1/2) × 10 = 5 V

Q. In the circuit shown in figure the maximum output voltage $V_{0}$ is

0 V

5 V

10 V

$\frac{5}{2} V$

For the positive half cycle of input the resulting network is shown below

$\Rightarrow (V_{0})_{max} = \frac{1}{2} (V_{i})_{max}$
$= \frac{1}{2} \times 10 = 5 V$

Q. In the circuit shown in figure the maximum output voltage V0​ is

0 V

5 V

10 V

2​5​V

For the positive half cycle of input the resulting network is shown below ⇒(V0​)max​=21​(Vi​)max​ =21​×10=5V

Q. In the circuit shown in figure the maximum output voltage $V_{0}$ is

$\frac{5}{2} V$

For the positive half cycle of input the resulting network is shown below

$\Rightarrow (V_{0})_{max} = \frac{1}{2} (V_{i})_{max}$
$= \frac{1}{2} \times 10 = 5 V$