JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 5)
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 :
_12th_April_Morning_Slot_en_5_1.png)
If V0 is almost zero, identify the correct statement :
If V0 is almost zero, identify the correct statement :
The value of the resistance R is 1.5 $$\Omega $$
The emf of the battery is l.5 V and its internal resistance is 1.5 $$\Omega $$
The emf of the battery is l.5 V and the value of R is 1.5 $$\Omega $$
The potential difference across the battery is 1.5 V when it sends a current of 1000 mA
Explanation
V = E – Ir
When V = V0 = 0 $$ \Rightarrow $$ 0 = E – Ir
$$ \therefore $$ E = r
When I = 0, V = E = 1.5 V
$$ \therefore $$ r = 1.5 $$\Omega $$
When V = V0 = 0 $$ \Rightarrow $$ 0 = E – Ir
$$ \therefore $$ E = r
When I = 0, V = E = 1.5 V
$$ \therefore $$ r = 1.5 $$\Omega $$
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