A) 40 W bulb will fuse
B) 100 W bulb will fuse
C) Both bulbs will fuse
D) Nothing will happen
Correct Answer: A
Solution :
From Joules law, power consumed by a bulb of resistance R is \[P=\frac{{{V}^{2}}}{R}\] where V is potential difference. Given, V= 200 volt, P=40W \[\therefore \] \[{{R}_{1}}=\frac{{{V}^{2}}}{P}=\frac{{{(200)}^{2}}}{40}=1000\,\Omega \] From Ohms law, \[V={{I}_{1}}R\] \[\therefore \] \[{{I}_{1}}=\frac{V}{R}=\frac{200}{1000}=\frac{1}{5}A\] Similarly, resistance of 100 W bulb, \[{{R}_{2}}=\frac{{{(200)}^{2}}}{100}\Omega =400\,\Omega \] \[\therefore \] \[{{I}_{2}}=\frac{1}{2}A\] when bulbs are connected in series, effective resistance \[R={{R}_{1}}+{{R}_{2}}\] \[=1000\,\Omega +400\,\Omega =1400\,\Omega \] When supply voltage is 440 V, then current \[I=\frac{440}{R}=\frac{440}{1400}=0.31\,A\] Since,\[I>{{I}_{1}}\]but less than hence the bulb of 40 W will fuse.
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