Question

Hey Mates !

May you please explain this (refer to attachment) in detail.
Please explain the underlined (red colored in 1st Page) text too.

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@ BeBrainliest :-)

Answer 1

Answer

Before solving get to know the below stuff,

• In a parallel connection the Voltage accross the circuit is the same and the current varies
• V = IR is the Ohm's law
• Here in the below Answer I've tried my best to simply the Answer and the steps might look a bit different still the Answer would be the same

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• Let's take a parallel connection of resistors say R₁ & R₂ then the net current in the circuit will be equal to the sum of the current that flows through R₁ & R₂ so,

➞ I = I₁ + I₂ ❲ -eq(1) ❳

• Also as we saw here at any point in the circuit the potential difference (voltage) will be the same so let's take it as V so for Resistor R₁ it would be,

➞ V = I₁R₁ ❲ -eq(2) ❳ [From the Ohm's law, V = IR]

Similarly for resistor R₂,

➞ V = I₂R₂ ❲ -eq(3) ❳

• Now from eq(2) & eq(3) we see that V (LHS) is common which means the RHS should give the same value,

➞ I₁R₁ = I₂R₂

➞ (I₁R₁)/R₂ = I₂ ❲ -eq(4) ❳

Or

➞ I₁ = (I₂R₂)/R₁ ❲ -eq(5) ❳

✭ Substituting eq(4) in eq(1) we get,

➞ I = I₁ + I₂

➞ I = I₁ + (I₁R₁)/R₂ [ Because (I₁R₁)/R₂ = I₂ ]

➞ I = I₁ ❲ 1 + R₁/R₂ ❳ [Taking I₁ common]

➞ I = I₁ ❲ (R₂ + R₁)/R₂ ❳

➞ I ❲ R₂/(R₂ + R₁) ❳ = I₁ ❲ -eq(6) ❳ [Taking I₂ to LHS]

• Similarly if we substitute eq(5) in eq(1) we get,

➞ I ❲ R₁/(R₂ + R₁) ❳ = I₂ ❲ -eq(7) ❳

✭ Dividing eq(6) by eq(7) we get,

➞ I₁/I₂ = R₂/R₁ [ Because R₂ + R₁ is common and gets cancelled ]

∴ The above Equations shows us that the current in each branch of a parallel connection is inversely proportional to the resistance in it