Physics, asked by Praveenraj11, 1 year ago

A current of 0.6 A is shown by ammeter in the circuit when the key K1 is closed. Find the

resistance of the lamp L. What change in current flowing through the 5 Ω resistor and potential

difference across the lamp will take place, if the key K2 is also closed. Give reason for your

answer.

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Answered by nandana9639

164

WHEN K2 is open we don't have to consider the 10Ω
I=0.6A
R=5Ω+RΩ=r
V=6V
r=10Ω
R=RESISTANCE oF Lamp=10-5
5Ω
there would be no change in the potential difference as in a parallel connection , it remains same

nandana9639: please mark as brainliest

nandana9639: thanks ;)

Praveenraj11: Thanks for your ans

Praveenraj11: Hi nanda, you have not explained about change in the current flowing.

Praveenraj11: can you answer that?

nandana9639: about current,
the total current flowing through the circuit=6/5=1.2A
this current is divided between two 10Ω in parallel connection
so,each part of the parallel connection gets 0.6A each and this is the current that flows through the the 5Ω resistor

nandana9639: hope this helps

Praveenraj11: Thank you, nanda

Answered by Adityaboy

Answer:

A current of 0.6 A is shown by ammeter in the circuit when the key K 1 is closed.

Find the resistance of the lamp L. What change in current flowing through the

5 Ω resistor and potential difference across the lamp will take place, if the key

K 2 is also closed. Give reason for your answer.

Explanation:

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