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In the circuit shown in figure potential difference between points $A$ and $B$ is $16 \mathrm{~V}$. the current passing through $2 \Omega$ resistance will be
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Verified Answer
The correct answer is:
$3.5 \mathrm{~A}$
$\therefore 4 \mathrm{i}_{1}+2\left(\mathrm{i}_{1}+\mathrm{i}_{2}\right)-3+4 \mathrm{i}_{1}=16 \mathrm{~V}$
Using Kirchhoff's second law in the closed loop we have
$$
9-i_{2}-2\left(i_{1}+i_{2}\right)=0
$$
Solving equations (i) and (ii), we get $\mathrm{i}_{1}=1.5 \mathrm{~A}$ and $\mathrm{i}_{2}=2 \mathrm{~A}$
$\therefore$ current through $2 \mathrm{~W}$ resistor $=2+1.5=3.5 \mathrm{~A}$
Using Kirchhoff's second law in the closed loop we have
$$
9-i_{2}-2\left(i_{1}+i_{2}\right)=0
$$
Solving equations (i) and (ii), we get $\mathrm{i}_{1}=1.5 \mathrm{~A}$ and $\mathrm{i}_{2}=2 \mathrm{~A}$
$\therefore$ current through $2 \mathrm{~W}$ resistor $=2+1.5=3.5 \mathrm{~A}$
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