A current of \(1\ ampere\) flows in a series circuit containing an electric lamp and a conductor of \(10\ \Omega\) when connected to a \(20\ V\) battery. Calculate the resistance of the electric lamp. Now if a resistance of \(15\) is connected in parallel with this series combination, what change (if any) in current flowing through \(10\ \Omega\) conductor? Give reason.
Let \(R_{1}\) be the resistance of electric lamp. Then, the total resistance is
By Ohm's law , we get the value \(R_1\) as \(\Omega\).
The resistance of the lamp is connected in series, then \(R\) \(=\) 20 \(\Omega\) .
Then, If it is connected with \(15\ \Omega\) resistance in parallel,
The value of current is given by the formula
On substituting the known values, we get, the current is \(A\)
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