ELECTRIC CIRCUITS

Answer :

I = {V / (5 + internal R of battery + Resistance of wire AB)}

                               = {6 / (5 + 1 + 2)}

                               = (3/4)A

                   emf of cell C = potential difference across AD

                                      = I × R

                                      = (3/4) × R

                   for 100 cm, R = 2Ω hence for AD = 60 cm, R = {(120) / (100)}Ω

                             ∴ emf of cell C = (3/4) × {(120) / (100)} = 0.9V  

Answer :

Resistance of 3Ω does not belong to any circuit where in a source  emf is connected.

                   ∴ There is no current through 3Ω resistor

                   Hence potential difference = I × 3

                                                          = 0 × 3

                                                          = 0 V

Answer :

When cell is connected to R₁, Let current be I. & internal resistance

                   be r.

                             hence I = {E / (R₁ + r)}

                             Heat generated = ɸ₁ = I² R₁ = {(E² R₁) / (R₁ + r)²}                   (1)

                   given is ɸ₁ is same as heat generated when connected to R₂. (say ɸ₂)

                             ɸ₂ = [{E²R₂} / {(R₂ + r)²}]                                                       (2)

                   As     ɸ₁ = ɸ₂

                             ∴ [{E²R₂} / {(R₂ + r)²}] = [{E²R₂} / {(R₂ + r)²}]

                             ∴ (R₁ / R₂) = {(R₁ + r) / (R₂ + r)}²

                             ∴ {(R₁ + r) / (R₂ + r)} = √(R₁ / R₂)

                             R₁√(R₂) + r√(R₂) = R₂√(R₁) + r√(R₁)

                             ∴ [{R₁√(R₂) – R₂√(R₁)} / {√(R₁) – √(R₂)}] = r

                             ∴ r = [{(R₁√(R₂) – R₂√(R₁)) (√R₁ + √R₂)} / {(√R₁ – √R₂) (√R₁ + √R₂)}]

                             = [{R₁√(R₁R₂) – R₁R₂ + R₁R₂} / {R₁ – R₂}] – R₂√R₁R₂

                             ∴ r = √R₁R₂

Answer :

It can be seen that, the resistive network can be drawn as

                   Hence it is balanced wheatstones bridge ∴ No current through 6R.

                   hence Reff = (R + 2R) || (4R + 2R) i.e. 3R || 6R

                             ∴ Reff = {(3R × 6R) / (3R + 6R)} = 2R

                   also I Total = {E / (Reff + R battery)} = {E / (4 + 2R)}

                             P = I_Total² (2R) = {E / (4 + 2R)}² × 2R

                             ∴ (dP / dR) = 2E² (d / dR) [R / {4 + 2R}²]

                             (dP / dR) = 2E²[1(4 + 2R)^–2 – 2(4 + 2R)^–3 ∙ 2R]

                   for P max (dP / dR) = 0 hence (4 + 2R)^–2 = 4R (4 + 2R)^–3

                             ∴ 1 = {(4R) / (4 + 2R)}

                   hence 4 + 2R = 4R

                             R = 2Ω   

Answer :

R = (ρℓ / A)                                                          (1)

                   after increasing length to 10 times, R¹ = 1000Ω

                                      ℓ' = 10ℓ.

                   volume = area × length = constant

                                      ∴ Aℓ = A'ℓ'

                                      ∴ (A' / A) = (ℓ / ℓ')

                   Now  R¹ = {(ρℓ') / A'}                                                             (2)

                   from (1) & (2)

                             ∴ (R / R') = (ℓ/A) × (A' / ℓ') = (ℓ / ℓ') × (ℓ / ℓ')

                             ∴ R = 1000 (ℓ / ℓ')² = 1000 × (1 / 10)² = 10Ω