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# How to solve question no. 4 ? See in image.

Anand Gokhale
18 Points
6 years ago
Given
Voltage = V = 2V
Resistance = R = 1 ohm
Time = t = 120 s
To find
no. of electrons flowing in circuit(since all electrons flow from negativeve to positive end only)= n = ?

Required Formulas
Ohm’s law V=IR
I= Q/t
Q=ne
V = voltage
I = current
R = resistance
Q = charge
t = time
n = no. of electrons
e = charge of 1 electron
Method
By Ohm’s law
V = IR
substituting for V and R
2V = I × 1 ohm
I = 1 A
now
I= Q/t
It=Q
t=120s
Q = 1C/s x 120s
q =120 C
q= ne
where n is no. of electrons and e is charge of one electron
n= q/e
n =120/(1.6 × 10-19)
n = 75 × 1019 electrons

Gaurav Manmode
22 Points
6 years ago
I= 2 A
noogler
489 Points
6 years ago
my ans is 15x1019
by ohms law …..........V=IR
2=ixR
I=2
as we knw that Q=ne,I=Q/t.............Q=charge,n=no. of electrns,e=charge of electrn,t=time in which charge flows
2=nx1.6x10-19/2x60
n=15x1019
Anand Gokhale
18 Points
6 years ago
sorry i made a mistake above since nobody else gave the right answer this is the real solution
Given
Voltage = V = 2V
Resistance = R = 1 ohm
Time = t = 120 s
To find
no. of electrons flowing in circuit(since all electrons flow from negativeve to positive end only)= n = ?

Required Formulas
Ohm’s law V=IR
I= Q/t
Q=ne
V = voltage
I = current
R = resistance
Q = charge
t = time
n = no. of electrons
e = charge of 1 electron
Method
By Ohm’s law
V = IR
substituting for V and R
2V = I × 1 ohm
I = 2 A
now
I= Q/t
It=Q
t=120s
Q = 2C/s x 120s
q =240 C
q= ne
where n is no. of electrons and e is charge of one electron
n= q/e
n =240/(1.6 × 10-19)
n = 150 × 1019 electrons = 1.5 X 1021electrons