Question

1. A 2Volt cell is connected to a 1Ω resistor. How may electrons come out of the negative terminal of the cell?

2. When two resistotrs R1 & R2 are connected in parllel, the net resistance is 3Ω. When connected in series, its value is 16Ω. Calaculate the value of R1 & R2.

Answer 1

1)
Voltage V= 2 volt;
Resistance R = 1Ω
Current I = V/R=2/1=2 Ampere = 2 coulomb/ sec
That is, 2×6.24×10¹⁸ =1.248×10¹⁸ electrons come out from the negative terminal of the cell every second.                   Ans.
And if the time t is given then the exact number of the electrons emitted will be 2 * t * 6.24 * 10¹⁸ electrons.
2)
Case 1:-- The net resistance of the parallel arrangement of the resistors R1 and R2 is Rp=3Ω  
 Case:--2 The net resistance of the series arrangement of the resistors R1and R2 is Rs=16Ω  ,  Rs=R1+R2
In case 2
Rs=R1+R2
 ⇒R1=16-R2 ---------------------------  equation 1
In case 1
1/Rp=1/R1+1/R2
Now substituting the value of the R1 from the equation 1 we get:--
1/3=1/(16-R2)+1/r2
For the solving the above Equation let R2 be x
Now
1/3=1/(16-x)+1/x
⇒1/3=16/(16x-x²)
⇒ 16x-x²=48
⇒x²-16x+48=0
⇒x²-12x-4x+48=0
⇒ x(x-12)-4(x-12)=0
⇒ (x-4)(x-12)=0
Now if x-12=0
then x=12 else if taking 
x-4=0
then x=4
So,R2=4Ω or 12Ω
Taking R2=12Ω
Then R1=16-R2=16-12=4Ω
and vice a versa.
Ans 2) Hence the two resistors are 4Ω and 12Ω.
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