JEE MAIN - Physics (2021 - 18th March Morning Shift - No. 18)
The voltage across the 10$$\Omega$$ resistor in the given circuit is x volt.
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The value of 'x' to the nearest integer is _________.
The value of 'x' to the nearest integer is _________.
คำตอบ
70
คำอธิบาย
_18th_March_Morning_Shift_en_18_2.png)
$${R_{eq}} = {{50 \times 20} \over {50 + 20}}$$
$${R_{eq}} = {{1000} \over {70}} = {{100} \over 7}\Omega $$
_18th_March_Morning_Shift_en_18_3.png)
$${R_{ef}} = 10 + {{100} \over 7} = {{170} \over 7}\Omega $$
_18th_March_Morning_Shift_en_18_4.png)
$$I = {V \over R} = {{170/7} \over {170}} = {{170} \over {170/7}}$$
I = 7A
Potential Across 10$$\Omega$$ resister
V10 = IR = 7 $$\times$$ 10 = 70V
$$ \Rightarrow $$ V10 = 70V