Question

(C) The peak voltage in a 220V, AC source is:-
(1) 220V
(2) 160V
(3) 310V
(4) 440V​

Answer 1

$\checkmark$ Given:

220 Volt AC source

$\checkmark$ To Find:

Peak voltage in a 220 Volt AC source

$\checkmark$ Solution:

We know that,

$\red{\bigstar \ \boxed{\rm V_{rms}=\dfrac{V_{0}}{\sqrt{2} }}}$

Here,

• V₀ = Peak Voltage

According to the Question,

We are asked to find the peak voltage in a 220 Volt AC source

So,

We must find the value of V₀

Therefore,

$\pink{\boxed{\boxed{\rm V_{0}=\sqrt{2} \times V_{rms}}}}$

Given that,

220 Volt AC source

Hence,

$\green{\rm \longrightarrow V_{rms}=220 \ V}$

Therefore,

Substituting the value,

We get,

$\red{\rm \implies V_{0}=\sqrt{2} \times 220 \ V}$

We know that,

$\blue{\longrightarrow \rm \sqrt{2}=1.414}$

Hence,

$\orange{\rm \implies V_{0}=1.414 \times 220 \ V}$

$\blue{\rm \implies V_{0}=311.08 \ V}$

$\blue{\rm \implies V_{0} \approx 310 \ V}$

Therefore,

Peak voltage in a 220 Volt AC source is 310 V

Hence,

$\checkmark$ Option (3) is correct

Answer 2

✰ Given -

• 220V in AC Source

✰ To Find -

• Peak Voltage in 220 source

✰ Solution -

We Know Formula for AC Source,

$✓ \: </strong><strong>V</strong><strong>(</strong><strong>R</strong><strong>ms) = \frac{</strong><strong>V</strong><strong>o}{ \sqrt{2} }$

Here,

• V(Rms) = Volt in AC Source
• Vo = Peak Voltage

According to Qn. Formula will Become,

$✓ \: Vo = \sqrt{2} \times v(Rms)$

Placing Values,

$⟶Vo = \sqrt{2} \times 220V$

$⟶Vo = 1.414 \times 220V$

$⟶Vo = 311.08V$

$⟶Vo = 310v$

⛬ Option (3) is Correct!