Question

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 1

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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Answer 2

$\mathtt{\huge{ \fbox{Solution :)}}}$

Given ,

The two 2 ohm and 3 ohm resistor are connected in parrallel combination with the battery of 2 volt

We know that , the equivalent resistance in parrallel combination is given by

$\mathtt{ \large \fbox{ \frac{1}{R} = \frac{1}{ R_{1}} + \frac{1}{R_{2}} + ... + \frac{1}{n} }}$

Thus ,

➡1/R = 1/2 + 1/3

➡1/R = (3 + 2)/6

➡1/R = 5/6

➡R = 6/5 ohm

We know that , the ohm's law is given by

$\huge \mathtt{ \fbox{V = IR} }$

Thus ,

➡2 = current × 6/5

➡current = 10/6

➡current = 1.67 amp

Hence , the 1.67 amp current passing through the entire circuit

Now , we have to find the current in individual resistor

Thus , in 2 ohm resistor

➡current = 2/2

➡current = 1 amp

And , in 3 ohm resistor

➡current = 2/3

➡current = 0.67 amp

Hence , 1 amp and 0.67 amp current are passing through the individual resistor

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