Question

Please solve this ??​

Answer 1

Explanation:

Let us solve this Problem ..

We are going to applied Kirchhoff,s law :

First in Loop AFEBA,

2I - I + 3I -6 = 0

Note : ( There is no current in BE of circuit )

So ;

5I = 7

I = 7 / 5 ............... ( i )

Now we are going to applied Kirchhoff,s Law in Loop ( A F D C A )

3I + RI - 4 - 6 + 2I - I = 0

5I + RI = 11 ................. ( ii )

Now , Substitute the value of I from equation ( i ) to the equation ( ii ).

Equation ( i ) is

I = 7 / 5

Equation ( ii ) is

5I + RI = 11

We get :

5 × 7/5 + R × 7/5 = 11

7 + 7R /5 = 11

7R / 5 = 4

R = 20 /7 ohm.

Now for potential difference across the point A and D along A F D ,

V(A) - 7 × 2 + 1 /5 - 3× 7 /5 = V(D).

V(A) - 14 / 5 + 1 - 21 /5 = V(D).

V(A) - V(D) = 14 / 5 + 21 / 5 - 1

V(A) - V(D) = 7 - 1 = 6 Volt.

Hence the Potential difference between the point A and D is ( 6 Volt ).

Thank you