First slide

Electrical power

Question

If a steady current I is flowing through a cylindrical element ABC. Choose the correct relationship :

Moderate

A

$V_{AB} = 2 V_{BC}$
B

Power across BC is 4 times the power across AB
C

Current densities in AB and BC are equal
D

Electric field due to current inside AB and BC are equal

Solution

$R_{B_{1}} = \frac{V^{2}}{100}, R_{B_{2}} = R_{B_{3}} = \frac{V^{2}}{60}$
Current in bulbs $I_{1} = I_{2} = \frac{250}{R_{B_{1} +} R_{B_{2}}} = \frac{250 x 300}{8 V^{2}}$
and $I_{3} = \frac{250}{R_{B_{3}}} = \frac{250 x 60}{V^{2}}$
Power in bulbs $W_{1} = {I_{1}}^{2} R_{B_{1}}, W_{2} = {I_{2}}^{2} R_{B_{2}}, W_{3} = {I_{3}}^{2} R_{B_{3}}$
$\Rightarrow W_{1} : W_{2} : W_{3} = {(\frac{250 x 300}{8 V^{2}})}^{2} (\frac{V^{2}}{100}) : (\frac{250 x 300}{8 V^{2}}) (\frac{V^{2}}{60}) : {(\frac{250 x 60}{V^{2}})}^{2} (\frac{V^{2}}{60})$
= 15 : 25 : 64

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