You are given three resistors of $10\, \Omega, 10\, \Omega, 20\, \Omega$,a battery of emf $2.5\, V$,a key,an ammeter,and a voltmeter. Draw a circuit diagram showing the correct connections of the given components such that the voltmeter gives a reading of $2.0\, V$.

(N/A) To get a reading of $2.0\, V$ across the $20\, \Omega$ resistor,we need to arrange the circuit such that the voltage divider rule provides the desired potential difference.
$1$. Connect the two $10\, \Omega$ resistors in parallel. Their equivalent resistance is $R_p = \frac{10 \times 10}{10 + 10} = 5\, \Omega$.
$2$. Connect this parallel combination $(5\, \Omega)$ in series with the $20\, \Omega$ resistor.
$3$. The total resistance of the circuit is $R_{eq} = 5\, \Omega + 20\, \Omega = 25\, \Omega$.
$4$. The total current in the circuit is $I = \frac{V}{R_{eq}} = \frac{2.5\, V}{25\, \Omega} = 0.1\, A$.
$5$. The voltage across the $20\, \Omega$ resistor is $V = I \times R = 0.1\, A \times 20\, \Omega = 2.0\, V$.
$6$. Thus,the voltmeter should be connected in parallel across the $20\, \Omega$ resistor as shown in the diagram.