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Standard XII

Physics

Mixed Combination of Cells

Two resistors...

Question

Two resistors, $R_{1}$ and $R_{2}$ , are identical, but the potential difference across $R_{1}$ is half the potential difference across $R_{2}$ . What is the ratio of the current in $R_{1}$ to the current in $R_{2}$ ?

\frac{1}{4}

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\frac{1}{2}

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1

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Solution

The correct option is B $\frac{1}{2}$
By Ohm's law, $V_{1} = I_{1} R_{1}$ and $V_{2} = I_{2} R_{2}$
Given, $V_{1} = \frac{V_{2}}{2}$ and $R_{1} = R_{2}$
So, $I_{1} R_{1} = \frac{I_{2} R_{1}}{2}$
Thus, $\frac{I_{1}}{I_{2}} = \frac{1}{2}$

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Two resistors, R1 and R2 , are identical, but the potential difference across R1 is half the potential difference across R2. What is the ratio of the current in R1 to the current in R2?

The correct option is B 12By Ohm's law, V1=I1R1 and V2=I2R2Given, V1=V22 and R1=R2So, I1R1=I2R12Thus, I1I2=12

Two resistors, $R_{1}$ and $R_{2}$ , are identical, but the potential difference across $R_{1}$ is half the potential difference across $R_{2}$ . What is the ratio of the current in $R_{1}$ to the current in $R_{2}$ ?

The correct option is B $\frac{1}{2}$
By Ohm's law, $V_{1} = I_{1} R_{1}$ and $V_{2} = I_{2} R_{2}$
Given, $V_{1} = \frac{V_{2}}{2}$ and $R_{1} = R_{2}$
So, $I_{1} R_{1} = \frac{I_{2} R_{1}}{2}$
Thus, $\frac{I_{1}}{I_{2}} = \frac{1}{2}$