Question 466: Current Electricity

Question

In the following circuit, the current through $6 \Omega$ resistor is

Accepted Answer

Applying Kirchhoff's second law for closed loop AEFBA we get,
$$ -\left(\mathrm{I}_1+\mathrm{I}_2ight) 	imes 6-\left(\mathrm{I}_1 	imes 3ight)+3=0 $$
$$ \Rightarrow \quad 9 \mathrm{I}_1+6 \mathrm{I}_2=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$$
Again, applying Kirchhoff's second law for a closed loop DEFCD we get,
$$ \begin{aligned} & -\left(I_1+I_2ight) 	imes 6-\left(I_2 	imes 3ight)+3=0 \ \Rightarrow & 6 I_1+9 I_2=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $$
Multiplying (i) by 2 and (ii) by 3 we get,
$$ 18 I_1+12 I_2=6 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$$
$$ 18 I_1+27 I_2=9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$$
Subtracting (iv) from (iii) we get,
$$ -15 I_2=-3 \Rightarrow I_2=\frac{1}{5} A $$
Substituting the value of $I_2$ in equation (i) we get,
$$ 9 I_1=3-6 	imes \frac{1}{5} \Rightarrow 9 I_1=\frac{9}{5} \Rightarrow I_1=\frac{1}{5} A $$
The current through the $6 \Omega$,
$$ =\mathrm{I}_1+\mathrm{I}_2=\frac{1}{5}+\frac{1}{5}=\frac{2}{5} \mathrm{~A} $$

Answer

$\frac{1}{5} \mathrm{~A}$

Answer

$\frac{2}{5} \mathrm{~A}$

Answer

$\frac{1}{4} \mathrm{~A}$

Answer

$\frac{3}{4} \mathrm{~A}$

Current Electricity

In the following circuit, the current through 6 \Omega resistor is

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