Question

A current of 1 ampere flows in series circuit contains an electric lamp and a conductor of 5 ohm when connected to a 10V battery. Calculate the resistance of the electric lamp.

Now if a resistance of 10 ohm is connected in parallel with this series combination, what change in current flowing through 5 ohm conductor and potential difference across the lamp will take place? Give reason.

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Answer 1

Part 1:
Let R be the resistance of the electric lamp. In series total resistance = 5 + R
I = $\frac{V}{R}$
1 =  $\frac{10}{5+R}$
R = 5 ohm 

Part 2:
Total Resistance =  $\frac{10 \times 10}{10+10} = 5\ ohm$
I =  $\frac{V}{R}=\frac{10}{5} = 2 Amp$
Current in each branch  = 2 / 2 = 1 Amp (since both branches have same resistances, current divides equally)

V across Lamp + conductor = 10 V
V acoess Lamp = I × R = 1 * 5 = 5 Volt
 

Answer 2

Answer:5v

Explanation:

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)

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