Filament of bulb

The filament of bulb has a 1 ohm resistivity and is connected to a voltage 220 V. How much electric charge will pass through the fiber when the electric current passes for 10 seconds?

Correct result:

Q =  2200 C

Solution:

U=220 V R=1 Ohm I=U/R=220/1=220 A t=10 s I=Q/t Q=I⋅ t=220⋅ 10=2200 CU=220 \ \text{V} \ \\ R=1 \ \text{Ohm} \ \\ I=U/R=220/1=220 \ \text{A} \ \\ t=10 \ \text{s} \ \\ I=Q/t \ \\ Q=I \cdot \ t=220 \cdot \ 10=2200 \ \text{C}

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