Answer:
The right solution is "4545.45 seconds".
Explanation:
The given values are:
Charge,
C = 2500 C
Resistance,
R = 200 Ω
Bulb plugged into,
V = 110 V
As we know,
⇒ [tex]I=\frac{V}{R}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{110}{200}[/tex]
⇒ [tex]=0.55 \ A[/tex]
Now,
⇒ [tex]Q=I\times t[/tex]
or,
⇒ [tex]t=\frac{Q}{I}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{2500}{0.55}[/tex]
⇒ [tex]=4545.45 \ seconds[/tex]
