Q13. The current flow in the circuit of two battery with internal resistance and two external resist

Q: What is the correct answer for CUET UG Physics Q13?

The correct answer is b. The circuit consists of two batteries (E₁ = 15 V, E₂ = 5 V) with internal resistances (r₁ = 1.2 Ω, r₂ = 0.8 Ω) and two external resistances (R₁_ext = 9 Ω, R₂_ext = 4 Ω) connected in series. The batteries are connected in opposition (their positive terminals face each other, or negative terminals face each other, implying they oppose each other's EMF). Net EMF = |E₁ - E₂| = |15 V - 5 V| = 10 V.Tota

Q: Q13. The current flow in the circuit of two battery with internal resistance and two external resistance as shown in figure will be (a) 1.5 A (b) 0.67 A (c) 0.9 A (d) 1.0 A

The correct answer is b. The circuit consists of two batteries (E₁ = 15 V, E₂ = 5 V) with internal resistances (r₁ = 1.2 Ω, r₂ = 0.8 Ω) and two external resistances (R₁_ext = 9 Ω, R₂_ext = 4 Ω) connected in series. The batteries are connected in opposition (their positive terminals face each other, or negative terminals face each other, implying they oppose each other's EMF). Net EMF = |E₁ - E₂| = |15 V - 5 V| = 10 V.Tota

Q: What are the options for this CUET UG Physics question?

A. 1.5 A · B. 0.67 A · C. 0.9 A · D. 1.0 A

Question

Q13. The current flow in the circuit of two battery with internal resistance and two external resistance as shown in figure will be
(a) 1.5 A
(b) 0.67 A
(c) 0.9 A
(d) 1.0 A

FirstInTest · Accepted Answer

The circuit consists of two batteries (E₁ = 15 V, E₂ = 5 V) with internal resistances (r₁ = 1.2 Ω, r₂ = 0.8 Ω) and two external resistances (R₁_ext = 9 Ω, R₂_ext = 4 Ω) connected in series. The batteries are connected in opposition (their positive terminals face each other, or negative terminals face each other, implying they oppose each other's EMF). Net EMF = |E₁ - E₂| = |15 V - 5 V| = 10 V.Total resistance in the circuit (R_total) = r₁ + R₁_ext + r₂ + R₂_ext = 1.2 Ω + 9 Ω + 0.8 Ω + 4 Ω = 15 Ω.Using Ohm's law, the current (I) flowing through the circuit is I = Net EMF / R_total = 10 V / 15 Ω ≈ 0.67 A.

Answer

1.5 A

Answer

0.67 A

Answer

0.9 A

Answer

1.0 A

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