Question

Two identical cells of emf 1.5V each joined in parallel, provide supply to an external circuit consisting of two resistors of 17 Ω. Each joined in parallel. A very high resistance voltmeter reads the terminal voltage of the cells to be 1.4V. What is the internal resistance of each cell ?

Answer 1

Let r is the internal resistance of cell , E is the emf of cell.
two cell of emf 1.5V and resistance r are in parallel also two resistors are joined in parallel with it as shown in figure.
Terminal voltage of the = 1.4V is given . Means voltage ( V) = 1.4V are given voltage between two resistance which are joined in parallel with circuit .

Now, we have to find out equivalent resistance,
first find equivalent resistance of two given resistors,
Req = R × R/(R + R) = R/2 = 17Ω/2 = 8.5Ω

now, solve the circuit by applying Kirchoff's law , Let i current flow through the circuit also i₁ and i₂ Current flow through each emf.
For first loop [ assume any loop in both ]
iReq + i₁r - E = 0
i₁ = (E - iReq)/r ----(1)
For second loop
iReq + i₂r - E = 0
i₂ = ( E - iReq)/r
now, I = i₁ + i₂ = (2E - 2iReq)/r
ir =2E - 2iReq
i = E/(Req + r/2)

Also V = iReq
I = V/Req = E/(Req + r/2)
Now, put V = 1.4V , E = 1.5V , Req = 8.5Ω
1.4/8.5 = 1.5/(8.5 + r/2)
1.4(8.5 + r/2) = 1.5 × 8.5
r = 1.21 Ω

Hence, answer is 1.21 Ω