JEE MAIN - Physics (2020 - 5th September Evening Slot - No. 11)
In the circuit, given in the figure currents in
different branches and value of one resistor are
shown. Then potential at point B with respect to
the point A is :
_5th_September_Evening_Slot_en_11_1.png)
+2 V
-2 V
+1 V
-1 V
Explanation
_5th_September_Evening_Slot_en_11_2.png)
Let us assume the potential at A = VA = 0
Now at junction C, According to KCL
i1 + i3 = i2
$$ \Rightarrow $$ 1 A + i3 = 2 A
$$ \Rightarrow $$ i3 = 1 A
Now Analyse potential along ACDB,
VA + 1 + i3(2) – 2 = VB
$$ \Rightarrow $$ 0 + 1 + 1(2) – 2 = VB
$$ \Rightarrow $$ VB = 3 – 2 = 1 V
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