Question

A $7V$ battery with internal resistance $2\Omega$. and a $3V$ battery with internal resistance $1\Omega$ are connected to a $10\Omega$ resistor as shown in figure,

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

Answer 1

Answer: (C)

Using Kirchhoffs law in loop A $P_2{P}_{1}DA$
$\therefore 10I_1+2I-7=0$
$10I_1+2I=7$
Using Kirchhoff's law in loop $P_2{P}_{1}CB{P}_2$
$-3+1\left(I-I_1\right)-10I=0$

$I-11I_1=3$;
$I=3+11I_1\dots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$,
$10I_1+2\left(3+11I_1\right)=7$;
$10I_1+6+22I_1=7$
$\therefore 32I_1=1$;
$I_1=1/32=0.031A$