Question

find the current in each arm of the given circuit .



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Answer 1

☀️ Here, in the figure, we are having three resistors with resistance 6Ω, 4Ω and 12Ω along with a cell of 1.5V potential difference.

☀️ We shall find the current flowing in each arm in the circuit.

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☀️ We are having two arms in the circuit namely :

• CE
• AB

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☀️ Firstly, we'll find the current between the first arm, AB :

☀️ There is only one resistor of 12Ω, so we can directly calculate the current using the formula :

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$\qquad: \implies~ \sf V= IR$

Where :

• V = Potential difference : 1.5V
• I = Current
• R = Resistance : 12Ω

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☀️ Finding the current :

$\sf \qquad : \implies \: 1.5V = (I) \: (12 \: ohms)$

$\qquad \sf : \implies \: I = \dfrac{1.5}{12}$

$\sf \qquad: \implies \: I = 0.125A$

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☀️ Therefore, the current in the first arm is 0.125A.

☀️ Now, we'll find the current in the second arm using the same formula.

☀️ Here, there are two resistors connected in series :

• CD = 6Ω
• DE = 4Ω

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☀️ Total resistance in CE = CD + DE

$\qquad \sf : \implies \: CD = 6 \: ohms + 4 \: ohms$

$\qquad \sf : \implies \: CD = 10 \: ohms$

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Here :

• V = Potential difference : 1.5V
• I = Current
• R = Resistance : 10Ω

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☀️ Finding the current :

$\qquad\sf : \implies \: V=IR$

$\sf\qquad : \implies \: 1.5V = (I) \: (10 \: ohms)$

$\sf \qquad : \implies \:I = \dfrac{1.5}{10}$

$\qquad \sf : \implies \: I = 0.15 A$

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☀️ Therefore, the current in the second arm is 0.15A.