Question

Find the current in the following circuit. ​

Answer 1

Concept Required :

• Series Combination - Always follow Only one path or direction.
• Rs - R1 + R2 + R3...Rn.

• In Parallel Combination - Always follow tow or more path or direction.
• 1/Rp - 1/R1 + 1/R2 + 1/R3 ...1/Rn.

Solution :

Let,

• Current be ( I )
• Resistance R.

$\rule{90}1$

A/Q,

• 3Ω and 7Ω are connected in series Combination,

$\tt \dagger \: \: \: \: \: Rs = R_1 + R_2 + R_3+...R_n$

Where as,

• Rs - Resistance connected in series Combination.
• R1 = 3Ω.
• R2 = 7Ω.

$\tt \longmapsto Rs = 7 + 3.$

$\tt \longmapsto Rs = 10 \Omega.$

Now

• 10Ω and 10Ω connected in parallel Combination.

$\tt\dagger \: \: \: \: \:\dfrac{1}{ R_p }= \dfrac{1}{R_1}+ \dfrac{1}{R_1}+\dfrac{1}{R_n}$

Here,

• R1 =10Ω.
• R2 = 10Ω.

$\tt\longmapsto\dfrac{1}{ R_p }= \dfrac{1}{10}+ \dfrac{1}{10}$

$\tt\longmapsto\dfrac{1}{ R_p }= \dfrac{2}{10}$

$\tt\longmapsto{ R_p }= {5} \Omega.$

And,

All Remaining Resistor Connected in Series Combination.

$\tt \longmapsto Rs = 7 + 3 + 5.$

$\tt \longmapsto Rs = 15 \Omega.$

Therefore, Total Equivalent Resistance is 15Ω.