ELECTRIC%20CIRCUITS

Answer :

For series connection, R = R₁ + R₂

                   ∴ R = {(ρ₁L) / A} + {(ρ₂L) / A} = (L/A) (ρ₁ + ρ₂)                     (1)

                   Where L is length of wire & A is cross-section area. Since the wires

                   are in series, combined length will be 2L, A will remain constant hence

                                      R = [{ρ(2L)} / A]                                                  (2)

                   from (1) & (2)

                                      {(2ρL) / A} = (L/A) (ρ₁ / ρ₂)

                                      ∴ ρ = {(ρ₁ + ρ₂) / 2}   

Answer :

Here  I₁ = (12 / 4) = 3A

                   Applying Kirchhoff’s law in ABCDA,

                             2I₂ + E – 4 I₁ = 0

                   given: I₂ = 0 & put I₁ = 3A hence

                             E = 4 I₁

                             E = 12V   

Answer :

This can be reduced as 1Ω series with 1Ω hence R = 2Ω

                   This 2Ω is parallel with 2Ω hence R' = (1/1)Ω

                   Now R' in series with 2Ω hence R" = 1 + 2 = 3Ω

                   R" parallel with 6Ω hence R"' = 3Ω || 6Ω

                                                              = 2Ω

                   This R"' & 2Ω in series. hence effective resistance is 2Ω + 2Ω = 4Ω

                   ∴ Total current = I = {(12V) / (4Ω)}

                             ∴ I = 3A

                   The circuit can be drawn as

                   at point C, I will get divide as (2I / 3) & (I/3).

                   further at point D, (2I / 3) will get divided equally as (I/3), (I/3)

                   ∴ current through ammeter = (I/3) = (3/3) = 1A   

Answer :

As deflection = 0, hence it’s balanced wheatstone bridge

                             ∴ (10 / 40) = (7/x)

                             ∴ x = 28Ω

Answer :

                   The circuit can be redrawn as shown.

                   Let a battery of EMF E is connected as shown.

                   The current in various branches is as shown.

                   applying kirchoff’s low to ACDA, CBDC, ADBGFA,

                             5i₂ + 5i₃ – 10i₁ = 0                                                                   (1)

                             10(i₂ – i₃) – 5(i₁ + i₃) – i₃ = 0                                          (2)

                             10 i₁ + 5(i₁ + i₃) – E                                                                 (3)

                   from (1) & (2) i₂ = {(3i₁) / 2} & i₃ = (i₁ / 2)

                   if R is effective resistance between A & B then

                             R = {E / (i₁ + i₂)} = {E / (i₁ + {(3i₁) / 2})} = {(2E) / (5i₁)}

                   from eqn (3), E = 15i₁ + 5i₃

                             ∴ E = 15i₁ + 5 × (i₁ / 2)

                                E = (35 / 2)i₁

                             ∴ R = [{2 × (35 / 2)i₁} / {5i₁}] = 7Ω