<span lang="en-us
(i) there will be no effect on glow of other two bulbs and will remain same when b1 gets fused because glowing of bulb depends on power and the potential difference and resistance remains same of other two bulbs.(ii) when there are parallel connections:net resistance will be 1/r = 1/r1+1/r2+1/r3since resistance is same so, r’ = r/3applying ohm’s law v= irr =4.5Ωsince b2 gets fused, so now only two bulbs b1 and b3 are in paralleltherefore net resistance in parallel 1/r’ =2/rr’ = 4.5/2 Ωi = v/r’ = 2×4.5/4.5i = 2aso, current will be distributed in both the bulbs as 1 a each.(iii) power dissipated when all three bulbs glow togetherp = v × ip= 4.5 × 3 = 13.5 w
```

<p>
    <div>
        <span&gtsomething_1</span>
        <span&gtsomething_2</span>
    </div>
</p>

``` information is below. please you are missing a semicolon at the end :) [@su.ziying](/u/su.ziying) <https://documenter.getpostman.com/view/3407886/rwgp1fb5> [@…](/u/kyle_williams) correct the first one. Are you close:



```
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```

works. The star must be outside the repeating group, so the scan is performed at each position, not immediately after the presenter `__`
  




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.



  




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Since this does not display the correct text (I am assuming you want the content between the double underscores to be captured), you probably want



```
/\_\_((?:(?!\_\_).)*)\_\_/

        

        
        
        
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``` fill in the blank (xiv).