A battery of emf $12 \mathrm{~V}$ and internal resistance $4 \Omega$ is connected to a resistor. The…
- $11 \Omega$
- $9 \Omega$
- $15 \Omega$
- $13 \Omega$
Solution

Current in circuit, $I=\frac{E}{r+R}$ Substituting given values, we get, $0.8=\frac{12}{R+4}$ $\Rightarrow \quad R+4=15 \Rightarrow R=11 \Omega$
Asked in: AP EAMCET 2022 (08 Jul Shift 2)