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# A torch bulb is rated $2.5V$and $750mA$. Calculate its power, its resistance, and the energy consumed if this bulb is lighted for $4hours$. Two identical resistors, each of resistance $2\Omega$ are connected in the torch in series and in parallel, to a battery of $12V$, calculate the ratio of power consumed in two cases.

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## Step 1: GivenTorch Voltage = $2.5V$Torch current = $750mA$Time = $4hours$Resistance = $2\Omega$Voltage of battery = $12V$ Step 2: Formula used and calculationPower is given by $P=VI\phantom{\rule{0ex}{0ex}}=2.5V×.75A\phantom{\rule{0ex}{0ex}}=1.875W$Power consumed by torch is $1.875W$.Resistance is given by$R=\frac{V}{I}\phantom{\rule{0ex}{0ex}}=\frac{2.5V}{0.75A}\phantom{\rule{0ex}{0ex}}=3.33\Omega$Resistance of torch is $3.33\Omega$.Energy consumed is given by$E=Pt\phantom{\rule{0ex}{0ex}}=1.875W×4×60×60s\phantom{\rule{0ex}{0ex}}=27000J$ In series combination${R}_{eq}={R}_{1}+{R}_{2}\phantom{\rule{0ex}{0ex}}=2\Omega +2\Omega \phantom{\rule{0ex}{0ex}}=4\Omega$$V=12V\phantom{\rule{0ex}{0ex}}I=\frac{V}{{R}_{eq}}\phantom{\rule{0ex}{0ex}}=\frac{12}{4}A\phantom{\rule{0ex}{0ex}}=3A\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{P}_{1}=VI\phantom{\rule{0ex}{0ex}}=12×3W\phantom{\rule{0ex}{0ex}}=36W$ In parallel combination$\frac{1}{{R}_{eq}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}+\frac{1}{2}\Omega \phantom{\rule{0ex}{0ex}}{R}_{eq}=1\Omega \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}V=12V\phantom{\rule{0ex}{0ex}}I=\frac{V}{{R}_{eq}}\phantom{\rule{0ex}{0ex}}=\frac{12}{1}A\phantom{\rule{0ex}{0ex}}=12A\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{P}_{2}=VI\phantom{\rule{0ex}{0ex}}=12×12W\phantom{\rule{0ex}{0ex}}=144W$ Ratio of power in series and parallel combination$\frac{{P}_{1}}{{P}_{2}}=\frac{36}{144}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}$Hence, the ratio is $1:4$.  Suggest Corrections  0      Similar questions  Explore more