Current Electricity Charging Discharging Of Capacitors Question 181
Question: In the adjoining circuit diagram each resistance is of 10 W. The current in the arm AD will be
Options:
A) $ \frac{2i}{5} $
B) $ \frac{3i}{5} $
C) $ \frac{4i}{5} $
D) $ \frac{i}{5} $
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Answer:
Correct Answer: A
Solution:
Applying Kirchoff’s law in mesh ABCDA $ -10(i-i _{1})-10i _{2}+20i _{1}=0 $
Therefore $ ,3i _{1}-i _{2}=i $ …….(i) and in mesh BEFCB $ -,20,(i-i _{1}-i _{2})+,10,(i _{1}+i _{2})+10i _{2}=0 $
Therefore $ 3i _{1}+4i _{2}=2i $ ……(ii) From equation (i) and (ii) $ i _{1}=\frac{2i}{5},,i _{2}=\frac{i}{5} $
Therefore $ i _{AD}=\frac{2i}{5} $
BETA
