Question 94: Current Electricity

Question

For the resistance network shown in the figure, choose the correct option(s).

Accepted Answer

Due to symmetry on upper side and lower side, points P and Q are at same potentials. Similarly, points S and T are at same potentials. Therefore, the simple circuit can be drawn as shown below :
$${I_2} = {{12} \over {2 + 2 + 2}} = 2A$$
$${I_3} = {{12} \over {4 + 4 + 4}} = 1A$$
$$	herefore$$ $${I_1} = {I_2} + {I_3} = 3A$$
I
PQ
= 0 because V
P
= V
Q
Potential drop (from left to right) across each resistance is
$${{12} \over 3} = 4V$$
$$	herefore$$ $${V_{MS}} = 2 	imes 4 = 8V$$
$${V_{NQ}} = 1 	imes 4 = 4V$$
or, $${V_S} < {V_Q}$$

Answer

The current through PQ is zero.

Answer

I1 = 3A.

Answer

The potential at S is less than that at Q.

Answer

I2 = 2A.

Current Electricity

For the resistance network shown in the figure, choose the correct option(s).

Choose the correct answer: