A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1 Ω. The current through the battery is

Question

A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1 Ω. The current through the battery is (A) 6 A (B) 8 A (C) 10 A (D) 12 A

Tardigrade Admin · Accepted Answer

Using symmetry and junction rule, we can arrange the currents as shown. Applying loop rule along ABCD and battery to A, we get

- i R - \frac{i}{2} R - i R + 10 = 0

As

R = 1Ω

∴ 10 = \frac{5 i R}{2} = \frac{5 i}{2}

or

i = \frac{20}{5} = 4 A

Current through the battery is

3 i = 12 A

.

Q. A battery of $10 V$ and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance $1Ω$ . The current through the battery is

$6 A$

$8 A$

$10 A$

$12 A$

Using symmetry and junction rule, we can arrange the currents as shown. Applying loop rule along ABCD and battery to A, we get
$- i R - \frac{i}{2} R - i R + 10 = 0$
As $R = 1Ω$
$∴ 10 = \frac{5 i R}{2} = \frac{5 i}{2}$
or $i = \frac{20}{5} = 4 A$
Current through the battery is $3 i = 12 A$ .

Q. A battery of 10V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1Ω. The current through the battery is

6A

8A

10A

12A