For a given a.c.,$\text{i} = \text{i}_{m}\sin\omega\text{t},$ show that the average power dissipated in a resistor R over a complete cycle is $\frac{1}{2}\text{i}_{m}^{2}\text{R}.$

A light bulb is rated at 100 W for a 220 V a.c. supply. Calculate the resistance of the bulb.

Q: 1. For a given a.c.,$\text{i} = \text{i}_{m}\sin\omega\text{t},$ show that the average power dissipated in a resistor R over a complete cycle is $\frac{1}{2}\text{i}_{m}^{2}\text{R}.$ 2. A light bulb is rated at 100 W for a 220 V a.c. supply. Calculate the resistance of the bulb.

1. From graph of $i^2 – t$ Average power consumed in resistor R $\text{P}_{av} = \frac{1}{\int_{0}^{T}\text{dt}}.\int_{0}^{T}\text{i}^{2}\text{R dt}$ $ = \frac{\text{i}_{m}^{2}\text{R}}{\text{T}}\int_{0}^{T}\sin^{2}\omega\text{t }\text{ dt} - - - - - (1)$ $ = \frac{\text{i}_{m}^{2}\text{R}}{2\text{T}}\int_{0}^{T}(1 -\cos2\omega\text{t})\text{dt}$ $ = \frac{\text{i}_{m}^{2}\text{R}}{2\text{T}}\bigg[\int_{0}^{T}\text{dt} - \int_{0}^{T}\cos2\omega\text{t }\text{ dt}\bigg] - - - - - (2)$ $ =\frac{\text{i}_{m}^{2}\text{R}}{2\text{T}}[\text{T} - 0 ]$ $ = \frac{\text{i}_{m}^{2}\text{R}}{2}$ 2. In case of ac $\text{P}_{av} =\frac{\text{V}_{rms}^{2}}{\text{R}} = \frac{\text{V}_{eff}^{2}}{\text{R}}$ $\Rightarrow\text{R} = \frac{\text{V}_{rms}^{2}}{\text{P}_{av}}$ $ = \frac{220\times220}{100}$ $ = 484\Omega.$

Question

For a given a.c.,$\text{i} = \text{i}_{m}\sin\omega\text{t},$ show that the average power dissipated in a resistor R over a complete cycle is $\frac{1}{2}\text{i}_{m}^{2}\text{R}.$
A light bulb is rated at 100 W for a 220 V a.c. supply. Calculate the resistance of the bulb.

CBSE OUTSIDE DELHI - SET 1 2013

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Solution

From graph of $i^2 – t$

Average power consumed in resistor R

$\text{P}_{av} = \frac{1}{\int_{0}^{T}\text{dt}}.\int_{0}^{T}\text{i}^{2}\text{R dt}$

$ = \frac{\text{i}_{m}^{2}\text{R}}{\text{T}}\int_{0}^{T}\sin^{2}\omega\text{t }\text{ dt} - - - - - (1)$

$ = \frac{\text{i}_{m}^{2}\text{R}}{2\text{T}}\int_{0}^{T}(1 -\cos2\omega\text{t})\text{dt}$

$ = \frac{\text{i}_{m}^{2}\text{R}}{2\text{T}}\bigg[\int_{0}^{T}\text{dt} - \int_{0}^{T}\cos2\omega\text{t }\text{ dt}\bigg] - - - - - (2)$

$ =\frac{\text{i}_{m}^{2}\text{R}}{2\text{T}}[\text{T} - 0 ]$
$ = \frac{\text{i}_{m}^{2}\text{R}}{2}$

In case of ac

$\text{P}_{av} =\frac{\text{V}_{rms}^{2}}{\text{R}} = \frac{\text{V}_{eff}^{2}}{\text{R}}$

$\Rightarrow\text{R} = \frac{\text{V}_{rms}^{2}}{\text{P}_{av}}$

$ = \frac{220\times220}{100}$

$ = 484\Omega.$

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