JEE MAIN - Physics (2024 - 5th April Morning Shift - No. 9)
In the given figure $$\mathrm{R}_1=10 \Omega, \mathrm{R}_2=8 \Omega, \mathrm{R}_3=4 \Omega$$ and $$\mathrm{R}_4=8 \Omega$$. Battery is ideal with emf $$12 \mathrm{~V}$$. Equivalent resistant of the circuit and current supplied by battery are respectively :
_5th_April_Morning_Shift_en_9_1.png)
$$12 \Omega$$ and $$11.4 \mathrm{~A}$$
$$10.5 \Omega$$ and $$1.14 \mathrm{~A}$$
$$10.5 \Omega$$ and $$1 \mathrm{~A}$$
$$12 \Omega$$ and $$1 \mathrm{~A}$$
Explanation
$$\begin{aligned}
R_{\mathrm{eq}}= & 12 \Omega \\
\text { and, } I & =\frac{E}{R_{\mathrm{eq}}} \\
& =\frac{12}{12} \\
& =1 \mathrm{~A}
\end{aligned}$$
Comments (0)
