ELECTRIC CIRCUITS

Answer :

                   Using detta of star conversion for portion ABD & BCE.

                   using formula, R₁ = {(RbRc) / (Ra + Rb + Rc)}

                   Value of each resistance will be {(1 × 1) / (1 + 1 + 1)} = (1/3)Ω

                   hence circuit can be drawn as

                   further reduction,

                   hence I = [4 / {(1/3) + (1/3) + {(2/3) || (5/3)}}] = [4 / {(2/3) + (10 / 21)}]

                               = [4 / (24 / 21)]

                           ∴ I = 4 × (21 / 24) = 3.5A    

Answer :

Converting left half delta to star

                   hence R will change to (R/3)

                   circuit can be drawn as

                   Hence equivalent resistance between A & B is Req =

                   Req = R || [(R/3) + {(4R / 3) || (4R / 3)}]

                           = R || [(R/3) + (2R / 3)]

                           = R || R

                   Req = (R/2)    

Answer :

effective R when in parallel is x

                   Hence

                             (1 / R₁) + (1 / R₂) + (1 / R₃) + --- {1 / (R_n–1)} + (1 / R_n) = (1/x) (1)

                   if n^th resistor is removed then.

                             (1 / R₁) + (1 / R₂) + ---- {1 / (R_n–1)} = (1/y)                               (2)

                             (1) – (2) gives

                             (1 / R_n) = (1/x) – (1/y)

                             ∴ (1 / R_n) = {(y – x) / (xy)} i.e. R_n = {(xy) / (y – x)}

Answer :

                   All 7 resistances are in parallel hence Reff = (70 / 7) = 10Ω

                             ∴ Total current = I = (14 / 10) = 1.4A.

                   circuit can be drawn as

                   current through (70 / 4)Ω be I₂.

                   hence I₂ = [{I × (70 / 3)} / {(70 / 3) + (70 / 4)}]

                             ∴ I₂ = 1.4 × [{1/3} / {(1/3) + (1/4)}]

                                   = 1.4 × [(1/3) / (7 / 12)]

                                   = 1.4 × [(1/3) × (12 / 7)]

                               I₂ = 0.8A

                   ∴ Ammeter will shown 0.8 A i.e. (4/5)A    

Answer :

Since the batteries in the upper part & lower part are connected in

                   reverse polarity, they will cancel out each other. Hence circuit can be

                   drawn as

                   Now remaining two cells are connected in reverse polarity.

                             Also internal resistance of ideal voltmeter = ∞

                   Hence resistance of battery across which it is connected will not get

                   change.

                             ∴ voltmeter reading = 5 – 5 = 0V