(a) Total resistance in the circuit  \(R=\frac { V }{ I } =\frac { 10V }{ A1 } =10\Omega \)
Since conductor of \(5\Omega \) and lamp are in series \(R=5\Omega +{ R }_{ lamp },{ \quad R }_{ lamp }=R-5\Omega =10\Omega -5\Omega =5\Omega \) Potential difference across lamp = \(I({ R }_{ lamp })=(IA)(5\Omega )=5V\)
(b) When \(10\Omega\)resistance is connected in parallel with \(R(=10\Omega )\)Total resistance (R') in the circuit is given by

    \(\frac { 1 }{ R' } =\frac { 1 }{ 10 } +\frac { 1 }{ R } =\frac { 1 }{ 10 } =\frac { 1 }{ 5 } \)
\(R'=5\Omega \)
Current through the circuit
\(I'=\frac { V }{ R' } =\frac { 10V }{ 5\Omega } =2A\)
Since \(10\Omega \) and R(=\(10\Omega \)) are in parallel current through R ie \(\frac { I' }{ 2 } =\frac { 2A }{ 2 } =1A\)
Thus current through lamp and conductor of \(5\Omega \)in series is 1A ie.
There is no change in current through conductor of \(​​​​5\Omega \).
Further potential difference across
\(lamp=\frac { I' }{ 2 } ({ R }_{ lamp })=1A(5\Omega )=5V\)

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