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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