An AC current is given by I = I₁sinωt + I₂cosωt. A hot wire ammeter will give a reading:

√((I₁² + I₂²)/2)

(I₁ + I₂) / 2√2

(I₁ + I₂) / √2

√((I₁² - I₂²)/2)

Ask & Answer Question

Alternating Current Overview Physics Alternating Current Physics Class 12th

An AC current is given by I = I₁sinωt + I₂cosωt. A hot wire ammeter will give a reading:

Option 1 -

(I₁ + I₂) / 2√2

Option 2 -

√((I₁² + I₂²)/2)

Option 3 -

(I₁ + I₂) / √2

Option 4 -

√((I₁² - I₂²)/2)

6 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

I = I? sin (ωt) + I? cos (ωt)

This can be written as I = I? sin (ωt + φ), where I? = √ (I? ² + I? ²)

A hot wire ammeter reads the rms value of the current.

I_rms = I? /√2 = √ (I? ² + I? ²)/√2

0 Upvotes 0 Downvotes

Similar Questions for you

Power factor of series LCR circuit at resonance is

Vishal Baghel

At resonance V_L = V_C

_{$\Rightarrow c o s ? = \frac{V_{R}}{\sqrt[]{V_{R}^{2} + {(V_{L} - V_{C})}^{2}}} = 1$}

An electric bulb and a capacitor are connected in series with an AC source. On increasing the frequency of the source, the brightness of the bulb:

Vishal Baghel

$X_{C} = \frac{1}{ω C}$

On increasing $ω$ XC decreases so Z decreases, current in circuit increases.

A direct current of 4 A and an alternating current of 4 A flows through two identical resistors. The ratio of heat produced in the two resistances in same time interval will be

Vishal Baghel

$\frac{H_{D . C .}}{H_{A . C .}} = \frac{I^{2} R}{I_{r m s}^{2} R} = 1$

The value of alternating emf (E) in the given circuit will be

Vishal Baghel

$E = \sqrt[]{V_{R}^{2} + {(V_{L} - V_{C})}^{2}}$

$E = \sqrt[]{{(60)}^{2} + {(100 - 20)}^{2}}$

E = 100V

Figure shows a circuit that contains four identical resistors with resistance R = $2.0 Ω,$ two identical inductors with inductance L = 2.0mH and an ideal battery with emf E = 9V. The current ‘i’ just after switch ‘S’ is closed will be: