Question

A battery having 12 V emf and internal resistance 3$T_{2}$ is connected to a resistor. If the current in the circuit is 1 A, then the resistance of resistor and lost voltage of the battery when circuit is closed will be

• A. 7$T_{1} > T_{2}$, 7 V
• B. 8$T_{1} < T_{2}$, 8 V
• C. 9$T_{1} = T_{2}$, 9 V
• D. 9$T_{1} = \frac{1}{T_{2}}$, 10 V

Answer 1

Answer: (C)

9$T_{1} = T_{2}$, 9 V

Answer 2

: Here, $\varepsilon = 12V,r = 3\Omega$and I= 1 A

$V = IR = \varepsilon - Ir$

$\therefore R = \frac{\varepsilon - Ir}{I} = \frac{12 - 1 \times 3}{1} = 12 - 3 = 9\Omega$

and , $V = IR = 1 \times 9 = 9V$