Question 311: Current Electricity

Question

The value of current in the 6 $$\Omega$$ resistance is :

Accepted Answer

The given figure can be represented as
From the above figure, it can be clearly seen that the voltage across point R is assumed as V.
Therefore, applying Kirchhoff's current law at point R, we can write
$${{V - 0} \over 6} + {{V - 90} \over 5} + {{V - 140} \over {20}} = 0$$
$$ \Rightarrow {V \over 6} + {{V - 90} \over 5} + {{V - 140} \over {20}} = 0$$
$$ \Rightarrow {{10V + 12V - 1080 + 3V - 420} \over {60}} = 0$$
$$ \Rightarrow 25V = 1500 \Rightarrow V = 60V$$
Therefore, current through 6$$\Omega$$ resistor is
$$I = {V \over R} = {{60} \over 6} = 10A$$

Answer

4A

Answer

8A

Answer

10A

Answer

6A

Current Electricity

The value of current in the 6 $\Omega$ resistance is :

Choose the correct answer: