JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Current Electricity

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To verify Ohm's law, a student connects the voltmeter across the battery as, shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained:

If

V_{0}

is almost zero, identify the correct statement:

(A)The potential difference across the battery is 1.5 V when it sends a current of 1000 mA .
(B)The emf of the battery is 1.5 V and the value of R is $1.5Ω$
(C)The emf of the battery is 1.5 V and its internal resistance is $1.5Ω$ ✓ Correct
(D)The value of the resistance R is $1.5Ω$

Answer: (C)

Solution

The terminal voltage of a cell follows

V = E - I r

, so a plot of

V

against

I

is a straight line whose intercept is the emf and whose slope magnitude is the internal resistance. From the graph, extrapolating to

I = 0

gives

V = E = 1.5 V

. At the other end, where

V_{0}

falls to about zero, the current is maximum and

0 = E - I_{m a x} r

, so

r = \frac{E}{I _{m a x}} = 1.5 Ω

. Thus the cell has an emf of

1.5 V

and an internal resistance of

1.5 Ω

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Current Electricity

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