Two cells of emf 1.5 V and 2 V and internal resistance 1Ω and 2Ω res

1.	Two cells of emf 1.5 V and 2 V and internal resistance 1Ω and 2Ω respectively are connected in parallel to pass a current in the same direction through an external resistance of 5Ω.(i) Draw the circuit diagram.(ii) Using Kirchhoff’s laws, calculate the current through each branch of the circuit and potential difference across 5Ω resistor.
Answer» (i) The circuit is shown in figure. (ii) Suppose I₁ are I₂ current drawn from cells ε₁and ε₂ respectively, then according to Kirchhoff’s junction law, current in R = 5Ω I = I₁ + I₂. Applying Kirchhoff’s second law to mesh ABFEA 1 × I₁ + 1.5 – 5(I₁ + I₂) = 0 ⇒ 6I₁ + 5I₂ = 1.5 …(i) Applying Kirchhoff’s second law to mesh CDEFC –2I₂ + 2 – 5(I₁ + I₂) = 0 ⇒ 5I₁ + 7I₂ = 2 …(ii) Solving equation (i) and (ii), we get \(I_1=\frac{1}{34}A,I_2=\frac{9}{34}A\) \(I=I_1+I_2=\frac{1}{34}+\frac{9}{34}=\frac{10}{34}A\) Potential difference across R = 5Ω resistor \((I_1+I_2)R=\frac{10}{34}\times5=\frac{25}{17}\) volt

Two cells of emf 1.5 V and 2 V and internal resistance 1Ω and 2Ω respectively are connected in parallel to pass a current in the same direction through an external resistance of 5Ω.(i) Draw the circuit diagram.(ii) Using Kirchhoff’s laws, calculate the current through each branch of the circuit and potential difference across 5Ω resistor.

Answer»

(i) The circuit is shown in figure.

(ii) Suppose I₁ are I₂ current drawn from cells ε₁and ε₂ respectively, then according to Kirchhoff’s junction law, current in R = 5Ω I = I₁ + I₂.

Applying Kirchhoff’s second law to mesh ABFEA

1 × I₁ + 1.5 – 5(I₁ + I₂) = 0

⇒ 6I₁ + 5I₂ = 1.5 …(i)

Applying Kirchhoff’s second law to mesh CDEFC

–2I₂ + 2 – 5(I₁ + I₂) = 0

⇒ 5I₁ + 7I₂ = 2 …(ii)

Solving equation (i) and (ii), we get

\(I_1=\frac{1}{34}A,I_2=\frac{9}{34}A\)

\(I=I_1+I_2=\frac{1}{34}+\frac{9}{34}=\frac{10}{34}A\)

Potential difference across R = 5Ω resistor

\((I_1+I_2)R=\frac{10}{34}\times5=\frac{25}{17}\) volt