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Home/Practice questions/Physics
PhysicsCurrent Electricity

For a resistor that obeys Ohm’s law, the graph of current I (y-axis)…

Case: Read the following passage and answer accordingly.

The current through a single resistor R across which there is a potential difference V is given by Ohm’s law \( I=V/R \). Resistors are sometimes joined together and there are simple rules for the calculation of equivalent resistance of such a combination. Two resistors are said to be in series if only one of their endpoints is joined. Two or more resistors are said to be in parallel if one end of all the resistors is joined together and similarly the other ends joined together.

For a resistor that obeys Ohm’s law, the graph of current \(I\) (y-axis) against potential difference \(V\) (x-axis) is a straight line. For two such resistors \(R_1\) and \(R_2\), the slopes of these lines are \(m_1\) and \(m_2\). The slope of the line for their series combination is

Answer & explanation+

Correct answer: option 4

The correct answer is Option (4) → \( \frac{m_1m_2}{m_1+m_2} \)

A resistor that obeys Ohm’s law has \( I=V/R \). So the slope of its \( I \)-\( V \) line is \( 1/R \). This gives \( R_1=1/m_1 \) and \( R_2=1/m_2 \).

In series the resistances add:

    \( \begin{aligned} R_s &= R_1+R_2 \\ &= \frac{1}{m_1}+\frac{1}{m_2} \\ &= \frac{m_1+m_2}{m_1m_2} \end{aligned} \)

The slope of the series line is the reciprocal of \( R_s \).

    \( \begin{aligned} m_s &= \frac{1}{R_s} \\ &= \frac{m_1m_2}{m_1+m_2} \end{aligned} \)

Therefore, the series line has the slope \( m_1m_2/(m_1+m_2) \). It is less steep than the line for either resistor alone.

NCERT (2026-27):
Ohm’s Law – “Imagine a conductor through which a current I is flowing and let V be the potential difference between the ends of the conductor. … V = R I … where the constant of proportionality R is called the resistance of the conductor.”

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