Question

A current of 1 ampere flows in a series circuit

containing an electric lamp and a conductor of 5 Ω

when connected to a 10 V battery. Calculate the

resistance of the electric lamp. Now if a resistance

of 10Ω is connected in parallel with this series

combination, what change (if any) in current flowing

through 5 Ω conductor and potential difference across

the lamp will take place? Give reason. Draw circuit

diagram.​

Answer 1

Explanation:

When 5Ω and resistance of lamp say R is connected in series:

Effective resistance R’ = 5+R

Applying ohm’s law:

V= IR

R = V/I = 10/1 = 10 Ω

Therefore, resistance of lamp, R= 10-5 Ω = 5 Ω

When a 10 Ω resistance is connected in parallel with the series connection

Effective resistance 1/R’’= 1/10 +1/10

R’’ = 5 Ω

Current through the circuit I = V/R = 10/5 = 2A

So, current flowing through each circuit will be of 1 V

Potential difference across the lamp will be same because voltage is same across the circuit.

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Answer 2

Explanation:

Total resistance of circuit can be calculated as follows:

$R=\frac{V}{I}$

$=\frac{10V}{1A}$

=10ohm

Since lamp and conductor are in series so resistance of lamp

10ohm-5ohm=5ohm

The new resistance in parallel to earlier combination has same value, i.e. 100 as the resistance of series combination. This means that the amount of current would be equally divided into two branches. Hence, 0.5A current will flow through 30 conductor

Now, resistance remains the same but current has become half. Using Ohm formula,

potential difference across the lamp can be calculated as follows:

V=IR=0.5A×1ohm=2.5V