AP EAPCET 22 May 2024 Shift 1 Solved Paper

Let the current in the loop be $I$.The upper branch has two $1\ \Omega$ resistors in series, so its resistance is $1+1=2\ \Omega$.The cells have emfs $E_2=12\ \mathrm{V}$ and $E_1=4\ \mathrm{V}$ opposing each other, so the net emf driving the current is $E_2-E_1=12-4=8\ \mathrm{V}$.The lower branch contains an $8\ \Omega$ resistor.Thus the total resistance of the loop is $2+8=10\ \Omega$.Using Ohm's law, $I=\frac{\text{net emf}}{\text{total resistance}}=\frac{8}{10}=0.8\ \mathrm{A}$.The potential difference between $P$ and $Q$ is the voltage across the $8\ \Omega$ resistor.So, $V_{PQ}=I\times 8=0.8\times 8=6.4\ \mathrm{V}$.Hence the current is $0.8\ \mathrm{A}$ and the potential difference between $P$ and $Q$ is $6.4\ \mathrm{V}$.Therefore, the correct option is $0.8\ \mathrm{A},\ 6.4\ \mathrm{V}$.