Question

The emf of a cell of negligible internal resistance is 2 V. It is connected to the series combination of $2\Omega ,3\Omega$ and $5\Omega$ resistances. The potential difference across $3\Omega$ resistance will be (in volt)

• A. $0.6$
• B. $\frac{2}{3}$
• C. $3$
• D. $6$

Answer 1

Answer: (A)

$0.6$

The emf of cell is, $E=2V$

The series equivalent of given resistances is

$R_s=2+3+5=10\Omega$

The current through the circuit is

$I=\frac{E}{R_s}$..............$($since, the cell has negligible internal resistance hence neglected$)$

$I=\frac{2}{10}=\frac{1}{5}A$

Now, the voltage across $3\Omega$ resistor is

$V=IR=\frac{1}{5}\left(3\right)$

$V=0.6V$