Current Electricity Charging Discharging Of Capacitors Question 148
Question: In the circuit shown below E1 = 4.0 V, R1= 2 W, E2 = 6.0 V, R2 = 4 W and R3 = 2 W. The current I1 is
[MP PET 2003]
Options:
A) 1.6 A
B) 1.8 A
C) 1.25 A
D) 1.0 A
Show Answer
Answer:
Correct Answer: B
Solution:
Applying Kirchhoff’s law for the loops (1) and (2) as shown in figure For loop (1) $ -2i _{1}-2(i _{1}-i _{2})+4=0 $
Therefore $ 2i _{1}-i _{2}=2 $ ?(i) For loop (2) $ -2(i _{1}-i _{2})+4i _{2}-6=0 $
Therefore $ -i _{1}+3i _{2}=3 $ ?(ii) On solving equation (i) and (ii) $ i _{1}=1.8A $ .
BETA
