Figure: http://img522. imageshack. us/img522/6987/yffigure26. . .

**Answer 1**

let a be the current in the upper branch { right to left}

And b be the current in the middle branch {LEFT TO RIGHT}

And then{ a – b} is the current in the lower branch.

Circuit containing upper and lower

2a +10a – 10b + 3a = 10

15 a – 10b = 10

3a – 2b = 2—————-1

Circuit containing middle and lower

– b + 10a – 10b – 4b = 5

10a – 15b = 5

2a – 3b = 1=======2

Solving a = 0.8A and b = 0.2 A

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Answers; 0.8A, 0.2A and 0.6 A

Vab = 3*0.8 – 4*0.2 = 1.6V

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Or

P.D across the upper branch = 10 – 5 a.

P.D across the middle branch = 5 – 5 b

P.D across the lower branch = 10*[a + b]

10 – 5 a. = 5 – 5 b = 10*[a + b]

—————————————-

Solving the equations we get the values

**Answer 2**

permit q = the fee, r = distance from the fee. Now the sphere variety the fee is in simple terms E = kq/r^2 the place ok = 9×10^9 N/(coul^2-m^2) capacity is V = -crucial(E*dl) the place dl is a vector alongside whose path you undertaking E, the “*” capacity dot or scalar product. For a element fee this works out uncomplicated as V = -crucial(E*dl)= – crucial(Edr) = -crucial(kq/r^2 dr) or V = -kq/r Now the aptitude distinction between 2 factors is a distinctive crucial: V(A-B) = -crucial(E*dl)from A to B. For a element fee this in simple terms turns into: V(A-B) = -kq*(a million/rB^2 -a million/rA^2) the place rA = decrease sure of crucial = commencing distance & rB = greater sure of crucial = end distance. For you subject: rA =0.25 m and rB = 0.seventy 5 m All you decide on now’s a calulator.

**Answer 3**

For part d: Vab = −(0.200 A)(4.00 Ω) − (0.800 A)(3.00 Ω) = −3.20 V.