JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 10)
_20th_July_Morning_Shift_en_10_1.png)
The value of current in the 6 $$\Omega$$ resistance is :
4A
8A
10A
6A
Explanation
The given figure can be represented as
_20th_July_Morning_Shift_en_10_2.png)
From the above figure, it can be clearly seen that the voltage across point R is assumed as V.
Therefore, applying Kirchhoff's current law at point R, we can write
$${{V - 0} \over 6} + {{V - 90} \over 5} + {{V - 140} \over {20}} = 0$$
$$ \Rightarrow {V \over 6} + {{V - 90} \over 5} + {{V - 140} \over {20}} = 0$$
$$ \Rightarrow {{10V + 12V - 1080 + 3V - 420} \over {60}} = 0$$
$$ \Rightarrow 25V = 1500 \Rightarrow V = 60V$$
Therefore, current through 6$$\Omega$$ resistor is
$$I = {V \over R} = {{60} \over 6} = 10A$$
From the above figure, it can be clearly seen that the voltage across point R is assumed as V.
Therefore, applying Kirchhoff's current law at point R, we can write
$${{V - 0} \over 6} + {{V - 90} \over 5} + {{V - 140} \over {20}} = 0$$
$$ \Rightarrow {V \over 6} + {{V - 90} \over 5} + {{V - 140} \over {20}} = 0$$
$$ \Rightarrow {{10V + 12V - 1080 + 3V - 420} \over {60}} = 0$$
$$ \Rightarrow 25V = 1500 \Rightarrow V = 60V$$
Therefore, current through 6$$\Omega$$ resistor is
$$I = {V \over R} = {{60} \over 6} = 10A$$
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