Question

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 1

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

Answer 2

Case 1

Given, Current = 1 A

Electric lamp and a conductor of 5 ohm are connected in series. Let resistance of electric lamp be R. So equivalent resistance would be R + 5 ohm. (We add resistance if it's a series circuit).

Potential difference of Battery = 10 V

We know that, V = IR

⇒ 10 = 1(R + 5)

⇒ 10 = R + 5

⇒ R = 5 ohm.

So, the resistance of electric lamp is 5 ohm.

Case 2

Resistance of 10 ohm is connected in parallel with the series combination.

Equivalent resistance of series = 5 + 5 = 10 ohm.

Now, equivalent resistance in whole circuit

$\sf \frac{1}{R_{eq}} = \frac{1}{10}+\frac{1}{10}\\\\\implies \frac{1}{R_{eq}} = \frac{2}{10}\\\\\implies R_{eq} = \frac{10}{2}\\\\\implies R_{eq} = 5 \ ohm$

Now, current flowing in circuit = I = V/R = 10/5 = 2 A

here, we are asked for current flowing in conductor.

Current flowing in conductor = current flowing in electric lamp = current flowing in series combination of lamp and conductor.

(In a series combination, current does not change)

So, I = V/R(eq)

⇒ I = 10/10 (Since, R(eq) = 10 ohm in series combination)

⇒ I = 1 A

The current across conductor doesn't change because another resistance of 10 ohm is added in parallel, which reduces the equivalent resistance and increases the current BUT divides it also. Hence, the current doesn't change.

Now, potential difference across lamp = V = IR

⇒ V = 1(5) (The resistance of lamp is 5)

⇒ V = 5 Volt

Previously, the potential difference in lamp was also 5 Volt, since the current was 1 A and the resistance was 5 ohm only. Hence, the potential difference also doesn't change because in a parallel combination, the potential difference does not change.