a current of 1A flows in a series circuit containg and electric lamp a

1.	a current of 1A flows in a series circuit containg and electric lamp and a conductor of 5ohm when conney to a 10V battery.Calculate the resistance of the electric lamp.Now if a resistance of 10ohm is connected in parallel with this series combination, what change(if any) in current flowing through 5ohm conductor and potential difference across the lamp will take place. give reason.
Answer» Here , In the Case 1 : V = 10 Total RESISTANCE in the series = 5+R Then, we know that I = V/R So, Here I = 10/(5+R) For Case 2: As the resistance are in parallel series so, Total Resistance = (R¹ × R² )/( R¹+R²) $= \frac{10 \times 10}{10 + 10} \\ \\ = 5 \: ohm$ Here, I = V/R I = 10/5 = 2 AMP So, current in each branch = 2/2 = 1 Reason - Because both of the BRANCHES have same resistance and because of it the current is divided equally. Potential across the lamp+ Conductor = 10V Potential across the lamp = 5 V

a current of 1A flows in a series circuit containg and electric lamp and a conductor of 5ohm when conney to a 10V battery.Calculate the resistance of the electric lamp.Now if a resistance of 10ohm is connected in parallel with this series combination, what change(if any) in current flowing through 5ohm conductor and potential difference across the lamp will take place. give reason.

Answer»

Here ,

In the Case 1 :

V = 10

Total RESISTANCE in the series = 5+R

Then, we know that I = V/R

So, Here I = 10/(5+R)

For Case 2:

As the resistance are in parallel series so,

Total Resistance = (R¹ × R² )/( R¹+R²)

$= \frac{10 \times 10}{10 + 10} \\ \\ = 5 \: ohm$

Here, I = V/R

I = 10/5 = 2 AMP

So, current in each branch = 2/2 = 1

Reason - Because both of the BRANCHES have same resistance and because of it the current is divided equally.

Potential across the lamp+ Conductor = 10V

Potential across the lamp = 5 V

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