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  • Fig. shows three similar lamps <img alt="\mathrm{L}_1, \mathrm{~L}_2 \text { and } \mathrm{L}_3" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BL%7D_1%2C%20%5Cmathrm%7B%7EL%7D_2%2

Fig. shows three similar lamps \mathrm{L}_1, \mathrm{~L}_2 \text { and } \mathrm{L}_3 connected across a power supply.

If the lamp \mathrm{L}_3 fuse, how will the light emitted by \mathrm{L}_1 \text{ and }\mathrm{L}_2 change ?

Option: 1

no change 


Option: 2

brilliance of \mathrm{L}_{1} decreases and that of \mathrm{L}_{2} increases


Option: 3

brilliance of both \mathrm{L}_{1} \text{ and }\mathrm{L}_{2} increases


Option: 4

brilliance of both \mathrm{L}_{1} \text{ and }\mathrm{L}_{2} decreases


Answers (1)

best_answer

Let R be the resistance of each lamp. If E be the applied e.m.f., then the current in the circuit, \mathrm{I_{1}}, is given by

\mathrm{I_1=\frac{E}{R+(R / 2)}=\frac{2 E}{3 R}}

Current flowing through \mathrm{L_{2}\: or\: } \mathrm{\mathrm{L}_3=\frac{1}{2}\left[\frac{2 \mathrm{E}}{3 \mathrm{R}}\right]=\frac{\mathrm{E}}{3 \mathrm{R}} }

When \mathrm{\mathrm{L}_3 } is fused, the whole current flows through \mathrm{\mathrm{L}_1 \text { and } \mathrm{L}_2 }.

Thus \mathrm{I_2=\frac{E}{R+R}=\frac{E}{2 R} }

Posted by

Riya

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