two resistors of resistance 2 ohm 3 ohm are connected in parallel to a

1.	two resistors of resistance 2 ohm 3 ohm are connected in parallel to a cell of EMF 2 volt and negligible internal resistance calculate the main current passing through the circuit and current through the individual resistance
Answer» Solution : ⏭ Given: ✏ Two resistors of resistance 2Ω and 3Ω are connected in parallel with p.d. of 2V ⏭ To Find: Net CURRENT passing through the circuit Current passing through both resistors ⏭ Formula: → Formula of equivalent resistance in parallel connection is given by... $\bigstar \: \boxed{ \pink{ \tt{ \large{ \dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}}}}} \: \bigstar$ → As PER OHM's law relation between POTENTIAL DIFFERENCE, current flow and resistance is given by... $\bigstar \: \boxed{ \tt{ \green{ \large{V = I \times R}}}} \: \bigstar$ ⏭ Calculation: _________________________________ Equivalent resistance $\implies \sf \: \dfrac{1}{R_{eq}} = \dfrac{1}{2} + \dfrac{1}{3} \\ \\ \implies \sf \: \frac{1}{R_{eq}} = \dfrac{3 + 2}{6} \\ \\ \implies \sf \: \boxed{ \tt{ \red{ \large{R_{eq} = \frac{6}{5} \: \Omega}}}} \: \orange{ \bigstar}$ _________________________________ Net current passing in the circuit $\mapsto \sf \: V = I_{net} \times R_{eq} \\ \\ \mapsto \sf \: 2 = I_{net} \times \dfrac{6}{5} \\ \\ \mapsto \sf \: I_{net} = \dfrac{5 \times 2}{6} \\ \\ \mapsto \sf \: I_{net} = \dfrac{10}{6} \\ \\ \mapsto \: \boxed{ \tt{ \blue{ \large{I_{net} = 1.67 \: A}}}} \: \orange{ \bigstar}$ _________________________________ Current in 2Ω resistor $\dashrightarrow \sf \: V = I \times R \\ \\ \dashrightarrow \sf \: 2 = I \times 2 \\ \\ \dashrightarrow \: \boxed{ \tt{ \purple{ \large{I = 1 \: A}}}} \: \orange{ \bigstar}$ _________________________________ Current in 3Ω resistor $\leadsto \sf \: V = I' \times R \\ \\ \leadsto \sf \: 2 = I' \times 3 \\ \\ \leadsto \boxed{ \tt{ \large{ \gray{I' = 0.67 \: A}}}} \: \orange{ \bigstar}$ _________________________________

two resistors of resistance 2 ohm 3 ohm are connected in parallel to a cell of EMF 2 volt and negligible internal resistance calculate the main current passing through the circuit and current through the individual resistance

Answer»

Solution :

⏭ Given:

✏ Two resistors of resistance 2Ω and 3Ω are connected in parallel with p.d. of 2V

⏭ To Find:

Net CURRENT passing through the circuit
Current passing through both resistors

⏭ Formula:

→ Formula of equivalent resistance in parallel connection is given by...

$\bigstar \: \boxed{ \pink{ \tt{ \large{ \dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}}}}} \: \bigstar$

→ As PER OHM's law relation between POTENTIAL DIFFERENCE, current flow and resistance is given by...

$\bigstar \: \boxed{ \tt{ \green{ \large{V = I \times R}}}} \: \bigstar$

⏭ Calculation:

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Equivalent resistance

$\implies \sf \: \dfrac{1}{R_{eq}} = \dfrac{1}{2} + \dfrac{1}{3} \\ \\ \implies \sf \: \frac{1}{R_{eq}} = \dfrac{3 + 2}{6} \\ \\ \implies \sf \: \boxed{ \tt{ \red{ \large{R_{eq} = \frac{6}{5} \: \Omega}}}} \: \orange{ \bigstar}$

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Net current passing in the circuit

$\mapsto \sf \: V = I_{net} \times R_{eq} \\ \\ \mapsto \sf \: 2 = I_{net} \times \dfrac{6}{5} \\ \\ \mapsto \sf \: I_{net} = \dfrac{5 \times 2}{6} \\ \\ \mapsto \sf \: I_{net} = \dfrac{10}{6} \\ \\ \mapsto \: \boxed{ \tt{ \blue{ \large{I_{net} = 1.67 \: A}}}} \: \orange{ \bigstar}$

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Current in 2Ω resistor

$\dashrightarrow \sf \: V = I \times R \\ \\ \dashrightarrow \sf \: 2 = I \times 2 \\ \\ \dashrightarrow \: \boxed{ \tt{ \purple{ \large{I = 1 \: A}}}} \: \orange{ \bigstar}$

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Current in 3Ω resistor

$\leadsto \sf \: V = I' \times R \\ \\ \leadsto \sf \: 2 = I' \times 3 \\ \\ \leadsto \boxed{ \tt{ \large{ \gray{I' = 0.67 \: A}}}} \: \orange{ \bigstar}$

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