Two resistors 4 Ω and 2 Ω are connected in series, and potential difference of 12 V is applied across the combination. Then, the current through the circuit is:

1 A

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2 A

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4 A

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4.5 A

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Solution

The correct option is B
2 A

Given:
resistance of two resistors = $R_{1} = 4 Ω a n d R_{2} = 2 Ω$
Equivalent resistance in series
$R = R_{1} + R_{2} = 4 Ω + 2 Ω = 6 Ω$

Potential difference $V = 12 v o l t$

Using Ohm's law: $I = \frac{V}{R} = \frac{12}{6} = 2 A$

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Two resistors 4 Ω and 2 Ω are connected in series, and potential difference of 12 V is applied across the combination. Then, the current through the circuit is:

The correct option is B 2 A Given: resistance of two resistors = R1=4 Ω and R2=2 Ω Equivalent resistance in series R=R1+R2=4 Ω+2 Ω=6 Ω Potential difference V = 12 volt Using Ohm's law: I=VR=126=2 A

The correct option is B
2 A

Given:
resistance of two resistors = $R_{1} = 4 Ω a n d R_{2} = 2 Ω$
Equivalent resistance in series
$R = R_{1} + R_{2} = 4 Ω + 2 Ω = 6 Ω$

Potential difference $V = 12 v o l t$

Using Ohm's law: $I = \frac{V}{R} = \frac{12}{6} = 2 A$