In the circuit shown below $E_{1} = 4.0\ V,\ R_{1} = 2\ \Omega,\ E_{2} = 6.0\ V,\ R_{2} = 4\ \Omega$ and $R_{3} = 2\ \Omega.$ The current $I_{1}$ is

Question

Dr. Abubacker · Accepted Answer

(b) Applying Kirchhoff’s law for the loops (1) and (2) as shown in figure
For loop (1)
$- 2i_{1} - 2\left( i_{1} - i_{2} \right) + 4 = 0$
$\Rightarrow 2i_{1} - i_{2} = 2$ …(i)
For loop (2)
$- 2\left( i_{1} - i_{2} \right) + 4i_{2} - 6 = 0$
$\Rightarrow - i_{1} + 3i_{2} = 3$ …(ii)
On solving equation (i) and (ii), $i_{1} = 1.8A$

Q. In the circuit shown below $E_{1} = 4.0\ V,\ R_{1} = 2\ \Omega,\ E_{2} = 6.0\ V,\ R_{2} = 4\ \Omega$ and $R_{3} = 2\ \Omega.$ The current $I_{1}$ is

Explanation

Reviewed by Dr. Abubacker

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