Question

Find the current through 10 ohm and 15 ohm when, I 1 = 1 Ampere​

Answer 1

Answer:

I(2) = 0.6 A

I(3) = 0.4 A

Explanation:

R(1)= 5 ohm

R(2)= 10 ohm

R(3)= 15 ohm

hence,

Equivalent Resistance Becomes :-

{ R(2) and R(3) in Parallel } In Series with R(1)

So,

$\frac{1}{r \frac{}{eq} } = \frac{1}{10} + \frac{1}{15} \\ \\ r \frac{}{eq} = \frac{15 \times 10}{25} = 6 \: ohm$

This R (eq) is in series with R(1)

Hence,

Equivalent Resistance of the entire circuit is equal to :-

$r = 5 + 6 = 11 \: ohm$

Now,

Current In Circuit = 1 Ampere

According to Ohm's Law :-

• V=IR

So,

Potential Difference of the circuit is :-

$v = ir \\ \\ v = 1 \times 11 = 11 \: volts$

Now,

Potential Drop across R(1) Resistance is :-

$v \frac{}{drop \: cross \: r1} = 1 \times 5 = 5volts$

Hence,

Potential Difference Between the resistors R(2) and R(3) each is :-

$11 - 5 = 6 \: volts$

Now For R(2) :-

$r = 10 \\ \\ v =6 \\ \\ i \frac{}{2} = \frac{v}{r} = \frac{6}{10} = 0.6 \: ampere$

Also we know :-

$i \frac{}{1} = i \frac{}{2} + i \frac{}{3} \\ \\ 0.6 + i \frac{}{3} = 1 \\ \\ i \frac{}{3} = 1 - 0.6 = 0.4 \: ampere$