A time varying voltage is given by -240sin120 t- where t is in second- This voltage is applied across a 120 - resistor- The average value of current for the half cycle is -

Given: ℰ=240sin120 t nbsp; and R=120 Ω So, ℰ_0=240 V Therefore, i_0=ℰ_0R=240120=2 A nbsp; We know that the average value of AC for a half cycle is given ...

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Standard XII

Physics

Resistors in Series and Parallel

A time varyin...

Question

A time varying voltage is given by $E = 240 sin 120 t$ , where $t$ is in second. This voltage is applied across a $120 Ω$ resistor. The average value of current for the half cycle is -

1.3 A

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1.9 A

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2.3 A

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2.9 A

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Solution

The correct option is A $1.3 A$
Given:

$E = 240 sin 120 t$ and $R = 120 Ω$

So, $E_{0} = 240 V$

Therefore, $i_{0} = \frac{E_{0}}{R} = \frac{240}{120} = 2 A$

We know that the average value of AC for a half cycle is given by,

$i_{a v g} = \frac{2 i_{0}}{π}$

$\Rightarrow i_{a v g} = \frac{2 \times 2}{π} = 1.3 A$

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A time varying voltage is given by E=240sin120t, where t is in second. This voltage is applied across a 120 Ω resistor. The average value of current for the half cycle is -

The correct option is A 1.3 AGiven: E=240sin120t and R=120 Ω So, E0=240 V Therefore, i0=E0R=240120=2 A We know that the average value of AC for a half cycle is given by, iavg=2i0π ⇒iavg=2×2π=1.3 A

A time varying voltage is given by $E = 240 sin 120 t$ , where $t$ is in second. This voltage is applied across a $120 Ω$ resistor. The average value of current for the half cycle is -