A65.0-ω resistor is connected to the terminals of a battery whose emf is 12.0 v and whose internal resistance is 0.5 ω. calculate (a) the current in the circuit, (b) the terminal voltage of the battery, vab, and (c) the power dissipated in the resistor r and in the battery’s internal resistance r.

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

= i R

= (0.1832) (65)

= 11.91 Volts

(c)

Power dissipated in the resister R is given as

= i²R

= (0.1832)²(65)

= 2.18 Watt

Power dissipated in the internal resistance is given as

= i²r

= (0.1832)²(0.5)

= 0.0168 Watt

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases . this process does not change the mass number.

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as -decay. in this the mass number remains same.

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

technology is the application of the knowledge gained by science.