1. A bulb drawn 0.4 A current from a 3 V battery in 6.5 minutes. Calculate the
(a) number of charge that flows in the circuit.
(b) resistance of the bulb filament.
(c) energy dissipated in the bulb.
Part (a)
I=QtI=\frac{Q}{t}I=tQ
0.4=Q60×6.50.4=\frac{Q}{60×6.5}0.4=60×6.5Q
Q=156CQ=156CQ=156C
Part (b)
V=IRV=IRV=IR
R=30.4=7.5ΩR=\frac{3}{0.4}=7.5\OmegaR=0.43=7.5Ω
Part (c)
E=QVE=QVE=QV
=156×3=156×3=156×3
=468J=468J=468J
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