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Mixed Combination of Cells

The resistanc...

Question

The resistance of Resistor A is 4 times greater than the resistance of Resistor B, if resistance A and B are connected in series then compare the following
1. Currents through these resistors ( $I_{A}$ and $I_{B}$ , respectively)
2. The voltage drops across them ( $V_{A}$ and $V_{B}$ , respectively)

I_{A} = I_{B}

and

V_{A} = V_{B}

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I_{A} = I_{B}

and

V_{A} = 4 V_{B}

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I_{A} = I_{B}

and

V_{B} = 4 V_{A}

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I_{A} = 4 I_{B}

and

V_{A} = 4 V_{B}

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I_{B} = 4 I_{A}

and

V_{B} = 4 V_{A}

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Solution

The correct option is A $I_{A} = I_{B}$ and $V_{A} = 4 V_{B}$
Here, $R_{A} = 4 R_{B}$
Since the resistances A and B are connected in series so the same current will pass through them. Thus, $I_{A} = I_{B}$
By Ohm's law, $V_{A} = I_{A} R_{A} = 4 I_{B} R_{B} = 4 V_{B}$

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The resistance of Resistor A is 4 times greater than the resistance of Resistor B, if resistance A and B are connected in series then compare the following 1. Currents through these resistors (IA and IB, respectively) 2. The voltage drops across them (VA and VB, respectively)

The correct option is A IA=IB and VA=4VBHere, RA=4RBSince the resistances A and B are connected in series so the same current will pass through them. Thus, IA=IBBy Ohm's law, VA=IARA=4IBRB=4VB

The correct option is A $I_{A} = I_{B}$ and $V_{A} = 4 V_{B}$
Here, $R_{A} = 4 R_{B}$
Since the resistances A and B are connected in series so the same current will pass through them. Thus, $I_{A} = I_{B}$
By Ohm's law, $V_{A} = I_{A} R_{A} = 4 I_{B} R_{B} = 4 V_{B}$