question_answer

What will happen when a 40 W, 220 V lamp and 100 W, 220 V lamp are connected in series across 40 V supply?

A)                  100 W lamp will fuse     

B)                  40 W lamp will fuse

C)                  Both lamps will fuse

D)                  Neither lamp will fuse

Correct Answer: B

Solution :

                When current i flows, across potential V, then power = \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\times 10\times {{(10)}^{2}}=500kg.m/s.\] and \[v=10m/s,\mu =0.5,\,m=10kg,\,g=10m/{{s}^{2}}\] (Ohms law)                 The currents required by the two lamps for their normal brightness are                                                                                                                                          \[s=\frac{\frac{1}{2}\times m\times {{(10)}^{2}}}{\mu mg}\]                 \[=\frac{50}{0.5\times 10}=10m.\]                 \[F=\mu R\] The resistance of the filaments are                 \[R=w=250N.\]                 \[\mu =0.3\]                 \[\therefore \]                 \[F=0.3\times 250=75N.\] The current in each lamp when connected in series with a 40 V supply is                 \[\overrightarrow{\mathbf{F}}\]                 \[\Delta t\]                 \[\overrightarrow{\mathbf{F}}\,\times \Delta t\]. Thus, \[\overrightarrow{\mathbf{F}}\,=m\,\overrightarrow{\mathbf{a}}\] and \[m=150g=150\times {{10}^{-3}}kg,a=20m/{{s}^{2}}\] Thus, the 40 W lamp will fuse, while the 100 W lamp will light dim. Alternative: Resistance of 40 W lamp is                 \[\therefore \] Resistance of 100 W lamp will be                 \[F=0.15\times 20=3N\] Now the current through series combination is given by                 \[=F\,\Delta t=3\times 0.1=0.3N.s.\] Potential drop across 40 W lamp will be given by                 \[{{v}^{2}}={{u}^{2}}-2gh\]                 \[\therefore \] and across 100 W lamp                 \[v=0\]                 \[\Rightarrow \] Therefore, 40 W bulb will fuse because the lamp can tolerate only 220 V.