Answer:[tex]121\ \Omega [/tex]
[tex]0.909\ A[/tex]
Explanation:
Given
Power [tex]P=100\ W[/tex]
Voltage applied [tex]V=110\ V[/tex]
Resistance of the bulb is given by
[tex]P=\frac{V^2}{R}[/tex]
[tex]100=\frac{110^2}{R}[/tex]
[tex]R=\frac{12100}{100}[/tex]
[tex]R=121\ \Omega [/tex]
Current drawn by the Power source is given by
[tex]P=V\cdot I[/tex]
[tex]I=\frac{P}{V}[/tex]
[tex]I=\frac{100}{110}[/tex]
[tex]I=0.909\ A[/tex]