Q.
A torch bulb rated as $4.5\, W , 1.5 \, V$ is connected as shown in fig. the e.m.f. of the cell, needed to make the bulb glow at full intensity is :
Current Electricity
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Solution:
Resistance of bulb is given as
$R_{b}=\frac{(1.5)^{2}}{4.5}=0.5 \, \Omega$
The equivalent of $0.5 \,\Omega$ and $1\, \Omega$ in parallel combination which is connected across the battery is given as
$R=\frac{0.5 \times 1}{0.5+1}=0.33 \,\Omega$
Current drawn from battery is given as
$i=\frac{E}{2.67+0.33}=\frac{E}{3}$
Current through the bulb is
$i_{ b }=\frac{E}{3} \times \frac{1}{1+0.5}=\frac{2 E}{9}$
For correct power dissipation in the bulb we use
$\left(\frac{2 E}{9}\right)^{2} \times 0.5=4.5$
$\Rightarrow E =13.5 \,V$
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