Chapter 17 - Electric Current and Resistance - Learning Path Questions and Exercises - Exercises - Page 619: 15

a). Terminal voltage of the battery = voltage drop across the resistance R = $IR=1.9A\times6 ohms=11.4 V$ So, terminal voltage = $11.4V$ b). $V=E-Ir=12-Ir$ or, $11.4=12-(1.9)r$ $r=0.32ohm$

a). Terminal voltage of the battery = voltage drop across the resistance R = $IR=1.9A\times6 ohms=11.4 V$ So, terminal voltage = $11.4V$ b). $V=E-Ir=12-Ir$ or, $11.4=12-(1.9)r$ $r=0.32ohm$