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Question

Physics Question on radiation

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 V0V_0 is almost zero, identify the correct statement:

A

The value of the resistance RR is 1.5Ω1.5 \,\Omega

B

The emf of the battery is 1.5V1.5\, V and the value of RR is 1.5Ω1.5 \, \Omega

C

The emf of the battery is 1.5V1.5\, V and its internal resistance is 1.51.5 Ω\Omega

D

The potential difference across the battery is 1.5V1.5\, V when it sends a current of 1000mA1000\, mA

Answer

The emf of the battery is 1.5V1.5\, V and its internal resistance is 1.51.5 Ω\Omega

Explanation

Solution

V=EIrV = E - Ir when V=V0=0    0=EIrV = V_0 = 0 \; \Rightarrow \; 0 = E - Ir   E=r\therefore \; E = r when I=0,V=E=1.5VI = 0, V = E = 1.5 \,V r=1.5Ω\therefore r = 1.5 \, \Omega.