Question

In the given circuit, an ideal voltmeter connected across the $10\Omega$ resistance reads $2V$ . The internal resistance $r$, of each cell is

• A. $1.5\Omega$
• B. $0.5\Omega$
• C. $1\Omega$
  
• D. $0\Omega$

Answer 1

The correct option is B $0.5\Omega$

Given the two cells are in series

Equivalent emf $V=1.5V+1.5V=3V$

Equivalent resistance of the circuit is
$R_{eq}=R_p+2\Omega +2r$
(series combination )

Where parallel combination

$R_p=\frac{15\Omega \times 10\Omega }{15\Omega +10\Omega }$

$R_p=6\Omega$

$R_{eq}=8+2r$

Let the current flowing in the circuit be $i$.
$\Rightarrow i=\frac{V}{Req}$

$i=\frac{3}{(8+2r)}$
(ohm’s law )

It is given that the voltage across $10\Omega$ is $2V$
(as measured by ideal voltmeter) this is same as voltage across the parallel combination $R_p$, current through $R_P$ is $i$

voltage across $10\Omega$ = voltage across $R_P$

$=i{R}_P$

$\Rightarrow i{R}_P=2$

Or

$(\frac{3}{(8+2r)}\times 6)=2$

$8+2r=9$ Or $r=0.5\Omega$

Hence option (D) is correct.