Question

A battery of e.m.f. 15 V and internal resistance 3 ohm is connected to two resistors of resistances 3 ohm and 6 ohm in series. The potential difference between the terminals of the battery is

• A. 12.5 V
• B. 9 V
• C. 11.25 V
• D. 10 V

Answer 1

The correct option is C 11.25 V

$\text{Given data:}$

$E=15V,r=3\Omega ,R_1=3\Omega .R_2=6\Omega$

$\text{Step 1: Draw circuit diagram and apply KVL [Refer Fig. 1]}$

While applying KVL in clockwise direction, increase in voltage is considered as positive and decrease in voltage as negative.

$E-i{R}_1-i{R}_2-ir=0$ .........$\left(1\right)$

$i=\frac{E}{R_1+R_2+r}$ ............$\left(2\right)$

$\text{Step 2: Finding terminal Potential Difference of Battery [Refer Fig. 2]}$

While writing potential difference $V_{AB}$, we go from B to A, increase in voltage is considered as positive and decrease in voltage as negative.

$V_{\text{terminal}}=V_{AB}=-ir+E=E-\frac{Er}{R_1+R_2+r}$

$=15-\frac{15\times 3}{3+6+3}=11.25V$

$V_{\text{terminal}}=11.25V$