NEET-XII-Physics
32: Electric Current in Conductors
- #38Find the current through the 10 Ω resistor shown in the figure (32-E14).
Figure 32-E14
Ans :
Applying KVL in loop 1, we get:
3i + 6i1 = 4.5 ...(1)
Applying KVL in loop 2, we get:
`` \left(i-{i}_{1}\right)10+3-6{i}_{1}=0``
`` 10i-16{i}_{1}=-3...\left(2\right)``
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Multiplying equation (1) by 10 and (2) by 3 and then, subtracting (2) from (1), we get:
`` -108{i}_{1}=-54``
`` \Rightarrow {i}_{1}=\frac{54}{108}=\frac{1}{2}=0.5``
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Substituting the value of i1 in (1), we get:
`` 3i+6\times \frac{1}{2}-4.5=0``
`` 3i-1.5=0``
`` \Rightarrow i=\frac{1.5}{3}=0.5``
So, current flowing through the 10 Ω resistor = i - i1 = 0.5 - 0.5 = 0 A
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