Question

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

• A. $1.5\Omega$
• B. $1\Omega$
• C. $1.2\Omega$
• D. $2\Omega$

Answer 1

Answer: (C)

$1.2\Omega$

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,

$Req=R\times R/(R+R)=R/2=17\Omega /2=8.5\Omega$

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 ]

$iReq+i₁r-E=0$

$i₁=\frac{(E-iReq)}{r}$ ----(1)

For second loop

$iReq+i₂r-E=0$

$i₂=(E-iReq)/r$

now, $I=i₁+i₂=(2E-2iReq)/r$

$ir=2E-2i{R}_{eq}$

$i=\frac{E}{(Req+r/2)}$

Also $V=iReq$

$I=V/Req=\frac{E}{(Req+r/2)}$

Now, put $V=1.4V,E=1.5V,R_{eq}=8.5\Omega$

$\frac{1.4}{8.5}=1.5/(8.5+r/2)$

$1.4(8.5+r/2)=1.5\times 8.5$

$r=1.21\Omega$

Hence, answer is $1.21\Omega$