Question

A 7 V battery with internal resistance 3$\left( e = 1.6 \times 10^{- 19}C \right)$ and a 3V battery with internal resistance 1$4\Omega$ are connected to a 10$2 \times 10^{- 2}\Omega$ resistor as shown in figure, the current in 10 $15\Omega$ resistor is

• A. 0.27 A
• B. 0.31 A
• C. 0.031A
• D. 0.53A

Answer 1

Answer: (C)

0.031A

Answer 2

: Using kirchhoff’s law loop A $P_{2}P_{1}$DA

$\therefore 10I_{1} + 2I - 7 = 0$

$10I_{1} + 2I = 7$ …(i)

Using kirchhoff’s law loop $P_{2}P_{1}CBP_{2}$

$- 3 + 1(I - I_{1}) - 10I_{1} = 0$

$I - 11I_{1} = 3$

$I = 3 + 11I_{1}$ …(ii)

From (i) and (ii)

$10I_{1} + 2(3 + 11I_{1}) = 7$

$10I_{1} + 6 + 22I_{1} = 7$

$\therefore$ $32I_{1} = 1$

$I_{1} = 1/32 = 0.031A$