Question

A new cell is marked 1.5V. When connected to an external resistance, the voltmeter connected to its terminal reads 1.1V. The decrease in potential across the terminals of the cell is due to

Answer 1

The decrease in potential across the terminals of the cell is due to the internal resistance of the cell.

Explanation:

→ The new cell is marked 1.5V. But when the cell is connected to an external resistance, the voltmeter connected to its terminal reads 1.1V.

→ This decrease in potential across the terminals of the cell is due to the internal resistance of the cell. Due to the internal resistance of the cell, the net resistance of the circuit increases. Hence the total potential of the cell gets divided between internal resistance and the external circuit.

→ Hence the effective electric potential available to the external circuit also decreases. Therefore when we connect a voltmeter across the terminal of the cell, we notice a reading smaller of 1.1 V reading of electric potential than the expected reading of 1.5 V.

Therefore when the new cell marked 1.5V is connected to an external resistance, the voltmeter connected to its terminal reads 1.1V due to the internal resistance of the cell.

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Answer 2

Answer: Due to internal resistance of the cell.

Explanation:

Voltage is the amount of effort needed to transfer a unit charge between two sites when a static electric field is present.

The potential difference between two points in a conservative vector field is referred to as potential difference.

When a cell is connected to an external source the external resistance is connected in series with the internal resistance of the cell. Due to which potential will drop both the resistances. The voltage drop across the external resistance is equal to the difference between internal voltage and potential drop to the internal resistance.

External resistance = R

Current in the circuit = i

Internal resistance = r

Cell potential = ∈

So, we can say that $\epsilon-ir=iR=V$= voltage drop across the terminal

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