16. Analyse the figure and answer the following questions.(a)If the sw

1.	16. Analyse the figure and answer the following questions.(a)If the switch S1, is on, which bulb will glow? How much current is flowing through the fuse at this time?b)If both the switches are on, what changes can be observed in the circuit ? Justify your answer.
Answer» Answer: Part a) Only BULB B1 will glow and the current through the switch is 0.87 A Part b) as soon as both switch is connected large current flow through fuse due to which it will melt and both bulbs will stop working Explanation: 1) When switch S1 is CLOSED then only Bulb 1 will glow as the circuit on left side is complete So here we can find the resistance of the bulb by using $R = \frac{V^2}{P}$ $R = \frac{230^2}{200}$ $R = 264.5 ohm$ now current through the fuse is given as $i = \frac{230}{264.5}$ $i = 0.87 A$ 2) When both switch is closed then we can see that both bulbs are identical and connected across the EMF source So here both bulb will withdraw same current from the cell So the total current from the fuse is $i = 0.87 + 0.87 = 1.74 A$ As we can see that fuse is connected with capacity of 1 A so as soon as both switch is connected large current flow through fuse due to which it will melt and both bulbs will stop working #Learn Topic : Ohm's law brainly.in/question/1589380

16. Analyse the figure and answer the following questions.(a)If the switch S1, is on, which bulb will glow? How much current is flowing through the fuse at this time?b)If both the switches are on, what changes can be observed in the circuit ? Justify your answer.

Answer»

Answer:

Explanation:

1) When switch S1 is CLOSED then only Bulb 1 will glow as the circuit on left side is complete

So here we can find the resistance of the bulb by using

$R = \frac{V^2}{P}$

$R = \frac{230^2}{200}$

$R = 264.5 ohm$

now current through the fuse is given as

$i = \frac{230}{264.5}$

$i = 0.87 A$

2) When both switch is closed then we can see that both bulbs are identical and connected across the EMF source

So here both bulb will withdraw same current from the cell

So the total current from the fuse is

$i = 0.87 + 0.87 = 1.74 A$

As we can see that fuse is connected with capacity of 1 A

so as soon as both switch is connected large current flow through fuse due to which it will melt and both bulbs will stop working

#Learn

Topic : Ohm's law