JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 24)
For the given circuit, in the steady state, $\left|\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{D}}\right|=$
________ V.
_31st_January_Evening_Shift_en_24_1.png)
Answer
1
Explanation
In steady state, capacitor behaves as an open circuit. Circuit is :
_31st_January_Evening_Shift_en_24_2.png)
$\Rightarrow i_{A B}=\frac{6}{3}=2 \mathrm{~A} $
$ i_{A D}=\frac{6}{12}=0.5 \mathrm{~A}$
$\Rightarrow V_{B}+2 \times 2-10 \times 0.5=V_{D}$
$\Rightarrow V_{B}-V_{D}=1$ volt
$\Rightarrow i_{A B}=\frac{6}{3}=2 \mathrm{~A} $
$ i_{A D}=\frac{6}{12}=0.5 \mathrm{~A}$
$\Rightarrow V_{B}+2 \times 2-10 \times 0.5=V_{D}$
$\Rightarrow V_{B}-V_{D}=1$ volt
Comments (0)
