A voltmeter of resistance `994Omega` is connected across a cell of emf 1 V and internal resistance `6Omega`. Find the potential difference across the voltmeter, that across the terminals of the cell and percantage error in the reading of the voltmeter.

A voltmeter of resistance 998 Omega is connected across a cell of emf 2 V and internal resistance 2 Omega . Find the p.d. across the voltmeter, that across the terminals of the cell and percentage error in the reading of the voltmeter.

(a) A voltmeter with resistance R_v is connected across the terminals of a battery of emf E and internal resistance r . Find the potential difference measured by the voltmeter. (b) If E = 7.50 V and r = 0.45 Omega , find the minimum value of the voltmeter resistance R_v so that the voltmeter reading is within 1.0% of the emf of the battery. (c) Explain why your answer in part (b) represents a minimum value.

A cell of e.m. E and internal resistance r is connected across a resistancer. The potential difference between the terminals of the cell must be

A resistance R is connected across a cell of emf epsilon and internal resistance r. A potentiometer now measures the potential difference between the terminals of the cell is V. Write the expression for r in terms of epsilon , V and R .

A voltmeter with resistance 500 Omega is used to measure the emf of a cell of internal resistance 4 Omega . The percentage error in the reading of the voltmeter will be

A cell whose e.m.f. is 2V and internal resistance is 0.1Omega is connected with a resistance of 3.9Omega the voltage across the cell terminal will be

A cell has an emf 1.5V. When connected across an external resistance of 2Omega , the terminal potential difference falls to 1.0V. The internal resistance of the cell is:

A cell which has an emf 1.5 V is connectedin series with an external resistance of 10Omega . If the potential difference across the cell is 1.25 V, then the internal resistance of the cell is ("in" Omega)

A cell of emf E is connected across a resistance R. The potential difference between the terminals of the cell is found to be V. The internal resistance of the cell must be -

The voltmeter shown in figure reads 18V across the 50(Omega) resistor. Find the resistance of the voltmeter.