Question

A 9.0V battery is connected to a bulb whose resistance is 2.4. How many electrons leave the battery per minute?

Answer 1

BY OHM'S LAW , V = IR , WHERE V IS THE POTENTIAL DIFFERENCE ,I IS THE CURRENT AND R IS THE RESISTANCE in ohms

   9 V = I * 2.4     

=> I = 3.75 A

1 AMPERE OF CURRENT IS 1 COULOMB /1 SECOND .

charge on 1 electron = 1.6 x 10^-19 COULOMB

Number of electrons in 1 coulomb =   1 / (1.6 X 10^-19) electrons

=  6.25 X 10¹⁸  ELECTRONS .

3.75 C OF CHARGE  LEAVES THE BATTERY IN 1 SEC .

THEREFORE, (3.75 X 60 ) C OF CHARGE LEAVES THE BATTERY IN 1 MINUTE .

So , (3.75 X 60 X 6.25 X 10^{18 )}  Electrons leaves the battery in one minute

THEREFORE 1.40625 X 10²¹ ELECTRONS   LEAVE THE BATTERY IN 1 MINUTE .