A voltmeter of resistance 998 Ω is connected across a cell of emf 2V and internal resistance 2 Ω . The potential difference across the voltmeter is:

Question

A voltmeter of resistance 998 Ω is connected across a cell of emf 2V and internal resistance 2 Ω . The potential difference across the voltmeter is: (A) 1.99V (B) 3.5 V (C) 5V (D) 6 V

Tardigrade Admin · Accepted Answer

From Ohms law i=(E/R+r)=(2/998+2)=2× 10-3A Potential difference across the voltmeter is V=iR=(2× 10-3)× 998=1.996V

Q. A voltmeter of resistance $998 Ω$ is connected across a cell of emf 2V and internal resistance $2 Ω$ . The potential difference across the voltmeter is:

1.99V

3.5 V

5V

6 V

1.5V

From Ohms law $i = \frac{E}{R + r} = \frac{2}{998 + 2} = 2 \times 10^{- 3} A$
Potential difference across the voltmeter is $V = i R = (2 \times 10^{- 3}) \times 998 = 1.996 V$

Q. A voltmeter of resistance 998Ω is connected across a cell of emf 2V and internal resistance 2Ω . The potential difference across the voltmeter is:

1.99V

3.5 V

5V

6 V

1.5V

From Ohms law i=R+rE​=998+22​=2×10−3A Potential difference across the voltmeter is V=iR=(2×10−3)×998=1.996V

Q. A voltmeter of resistance $998 Ω$ is connected across a cell of emf 2V and internal resistance $2 Ω$ . The potential difference across the voltmeter is:

From Ohms law $i = \frac{E}{R + r} = \frac{2}{998 + 2} = 2 \times 10^{- 3} A$
Potential difference across the voltmeter is $V = i R = (2 \times 10^{- 3}) \times 998 = 1.996 V$