Question

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

Answer 1

Let us consider

R as the resistance of lamp

Hence

Series resistance = 5 + R

We know that

Ohms law

I = V/R

Here

• I = current
• V = voltage
• R = resistance

Hence

1 = 10/5+R

R = 5 ohm

Therefore

Resistance of Lamp = 5 Ω

Now

Calculating total resistance in Parallel connection

Therefore

Total resistance = 10 × 10/10 + 10

On further simplification

We get

Total resistance = 5 ohm

Hence

Again using Ohms Law

We get

I = V/R

I = 10/5

I = 2 Ampere

Therefore

Current = 2 A