How Much Energy Does a Lamp Dissipate with a 0.3 A Current and 6 V Power Supply?

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A lamp with a current of 0.3 A and a 6 V power supply dissipates electrical energy over two minutes. The calculations show that the total charge (Q) is 36 coulombs, leading to an energy dissipation of 216 J using the formula E = QV. The initial approach of calculating power as I*V and then multiplying by time is also mentioned as a valid method. The discussion emphasizes the importance of understanding the relationship between current, voltage, and energy dissipation. Overall, the calculations confirm that the lamp dissipates 216 J of energy.
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A current of 0.3 A is passed through a lamp for 2 minutes using a 6 V power supply. The electrical energy dissipated by this lamp during the two minutes is:
a) 1.8 J
b) 12 J
c) 20J
d) 36 J
e) 216 J

If someone could just help me to start off that would be great.
The main equations I have to use are

I=Q/t
E= QV

(.3A)= (Q)/(120)
Q= 36

E= (36)(6V)
E=216 J

Am I right:confused:
 
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looks good, tho ordinarily the power/heat would first be calculated as I*V, then multiplied by 120
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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