NCERT Exemplar XXVI — task. Science CBSE, Class 10.

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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 \(10\)

Ω

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.

Let \(R_{1}\) be the resistance of electric lamp. Then, the total resistance is

R = i_{i} + i

By Ohm's law , we get the value \(R_1\) as

Ω

The resistance of the lamp is connected in series, then \(R\) \(=\)

Ω

. Then,

\begin{array}{l} \frac{1}{R_{P}} = \frac{i}{i} \\ R = i Ω \end{array}

The value of current is given by the formula

I = \frac{i}{i}

On substituting the known values, we get \(A\).

Did you find an error?

85. NCERT Exemplar XXVI