Question

A small toy car draws a 0.50mA from a 3.0 V nickel-cadmium(NiCd) battery. In 10 min of operation, (a) how much charge flows through the toys car? (b) How much energy is lost by the battery?

Answer 1

Given that

Current(I) = 0.50mA

                 = 0.50*10^-3A

emf of the batery(?) = 3V

Tiem(t) = 10min

             = 600sec

Since we know the formula for from the Ohm's law is

              V = IR

              R = V/I

                  = (3V)/(0.50*10^-3A)

                  = 6*10³?

Now the amount of energy lost is

           E = I²Rt

              =(0.50*10^-3A)²(6*10³?)(600sec)

              = 0.9J

Charge(q) = It

                = (0.50*10^-3A)(600sec)

                = 0.3C