Question

derivation of resistors in series and in parallel ?

Answer 1

$\huge{ \green{ \mathfrak{ \underline{Answer}}}}$

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$\boxed{\red{\bold{Resistors\: in\: series:}}}$$<font color="2">$

For resistors in series, each resistor has same current flowing through them but have different potential difference , and the sum of them is the potential difference supplied by the battery.

Let the resistors be $R_1 , R_2 , R_3$ and their potential difference be $V_1 , V_2 , V_3$ respectively. I is the current supplied to them.$</font>$

$\mathcal{\blue{Then:}}$

$V = V_1 + V_2 + V_3$

$\boxed{\red{\bold{By \:Ohm's\: Law:}}}$

$V = IR$

$V_1 = IR_1$

$V_2 = IR_2$

$V_3=IR_3$

$\boxed{\red{\bold{Plugging\: in \:values:}}}$

$IR = IR_1 + IR_2 + IR_3$

$IR = I(R_1 + R_2 + R_3)$

$R = R_1 + R_2 + R_3$

$\boxed{\red{\bold{Resistors \: in\:Parallel:}}}$$<font size="3">$

For resistors in parallel, each resistor has same potential difference through them but have different current flowing through them, and the sum of them is the current supplied in the circuit.

For resistors in parallel, each resistor has same potential difference through them but have different current flowing through them, and the sum of them is the current supplied in the circuit.Let the resistors be$R_1 , R_2 , R_3$ and the current through them be $I_1 , I_2 , I_3$ respectively. V is the potential difference supplied to them.$</font>$

$\mathcal{\blue{Then:}}$

$I = I_1 + I_2 + I_3$

$\boxed{\red{\bold{By\: Ohm's \:Law:}}}$

$I = \displaystyle{\frac{V}{R}}$

$I_1 = \displaystyle{\frac{V}{R_1}}$

$I_2 = \displaystyle{\frac{V}{R_2}}$

$I_3 = \displaystyle{\frac{V}{R_3}}$

$\boxed{\red{\bold{Plugging\:in\:values:}}}$

$\displaystyle{\frac{V}{R}} = {\frac{V}{R_1}}+ {\frac{V}{R_2}} + {\frac{V}{R_3}}$

$\displaystyle{\frac{V}{R}} =V ×\left( {\frac{1}{R_1}} + {\frac{1}{R_2}} + {\frac{1}{R_3}} \right)$

${\displaystyle{\frac{1}{R}} = {\frac{1}{R_1}} + {\frac{1}{R_2}} + {\frac{1}{R_3}}}$

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