Question

find equivalent resistance and total current
plz answer quickly
correct answer will be marked as brainliest

Answer 1

Explanation:

• All 6 ohm resistors are in parallel so there net resistance is

$\frac{1}{r} = \frac{1}{r1} + \frac{1}{r2} + \frac{1}{r3}$

$\frac{1}{r} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6}$

$\frac{1}{r} = \frac{3}{6}$

$\frac{1}{r} = \frac{1}{2}$

$r = 2 \: ohm$

• Now 12 ohm resistors are also in parallel, there net resistance is

$\frac{1}{r} = \frac{1}{r1} + \frac{1}{r2}$

$\frac{1}{r} = \frac{1}{12} + \frac{1}{12}$

$\frac{1}{r} = \frac{2}{12}$

$\frac{1}{r} = \frac{1}{6}$

$r = 12 \: ohm$

• Now net resistance of all 6 ohm resistors, net of 12 ohm resistors and 3 ohm resistance are in series
• So, Net resistance of circuit is

$r = r1 + r2 + r3$

$r = 2 + 3 + 6$

$r = 11 \: ohm$

• Current in circuit is

$current = \frac{emf \: of \: battery}{net \: resistnce}$

$i = \frac{30}{11}$

$i = 2.72 \: ampere$