Question

Solve the numerical ques 6 in the attachment

Standard:- 10

Content Quality Solution Required

❎ Don't Spamming ❎

Answer 1

Your answer is --

when K1 key is closed then , the reading of ammeter is 0.6A and the potential difference of hole circuit is 6V.

In seres combination Current is same buy potential difference is different for all resistance .

So, by ohm's law potential difference acceso 5ohm resistance is

V = 5 × 0.6

=> V = 3Volt

So, potential different across bulb is
6 -3 = 3volt .

Now, according to ohms law resistance of lamp is

R = 3/0.6

=> R = 5ohm .

Now, we find the resultant resistance of 5ohm resistor and lamp

So, equivalent resistance = 5+5 = 10ohm

Now, when K2 key is open then the current also start flowering in the above part of circuit and in above part of circuit 10ohm resistance is attached in parallel combination .

So, current is distributed in both resistance 10ohm and equivalent resistance of lamp and 5ohm resistor (10ohm) respectively but potential different is same because we know that potential difference is same in parallel combination .

Now, according to ohms

current in 10ohm resistance is

l = 6/10

= 0.6A

it means current is same across 5ohm resistance .

Now, 0.6A current flow across 5ohm resistance and lamp is 0.6A and reactance of lamp is 5ohm .

it means potential difference across lamp is also same.


【 Hope it helps you 】

Answer 2

● When only key k1 is closed then current will start flowing in the branch.

◢THEN

=> Resistance = 5 + L

=> V = 6v

=> I = 0.6A

=> R = 5 + L

=> V = IR

=> R = V/I

=> 5 + L = 6/0.6 = 10

=> L = 5 ohm

_________________________

● Potential difference across Lamp = 5×0.6 = 3v

● Now when key k2 is closed then current will start flowing in top most branch also.

● Current will be divided in two half as both the Resistance will be equal to 10ohm of each branch.

● NOW, Current through 5 ohm resistor = 0.3 A

● Potential difference across lamp = 0.3×5 = 1.5v

=======================================

____☆$BE \: \: \: \: BRAINLY$☆____